Class 9 Maths Ganita Manjari Chapter 4 Solutions — Exploring Algebraic Identities

Q: What is a3 + b3 + c3 − 3abc?

It equals (a + b + c)(a2 + b2 + c2 − ab − bc − ca). If a + b + c = 0, then a3 + b3 + c3 = 3abc.

These Class 9 Maths Ganita Manjari Chapter 4 solutions cover Exploring Algebraic Identities from the new NCF-2023 textbook (2026–27). Checked against NCERT Ganita Manjari, Grade 9, Part I, first edition April 2026, pages 68–91. The guide follows Exercise Sets 4.1–4.5 and the 13 end-of-chapter questions. Restrictions on denominators and positive dimensions are retained; textbook inconsistencies are explicitly flagged.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 4 Exercises: 4.1–4.5, End-of-Chapter Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
All identities (formulas)
Exercise Set 4.1
Exercise Set 4.2
Exercise Set 4.3
Exercise Set 4.4
Exercise Set 4.5
End-of-Chapter Exercises (Q1–Q13)
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 4 Overview

Chapter 4 of Ganita Manjari, Exploring Algebraic Identities, develops the standard algebraic identities and shows three uses for each: fast numerical calculation, expanding expressions, and factorising them (including with algebra tiles). It then extends to the cubic identities and to simplifying rational expressions by cancelling common factors. Use the labelled area models alongside the algebra, and check a factorisation by expanding it back.

Key Concepts & Definitions

Identity: an equation that is true for all values of the variables (e.g. $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ ).

Expanding: using an identity to remove brackets.

Factorising: writing an expression as a product of factors, often by recognising an identity or by splitting the middle term.

Splitting the middle term: for $x^{2}+\left(a+b\right)x+ab$ , find a and b with sum = coefficient of x and product = constant.

Rational expression: a ratio of two polynomials; simplify it by factorising and cancelling common factors (denominator ≠ 0).

All Identities Used in This Chapter

$\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$

$\left(a+b+c\right)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

$a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)$ $\left(x+a\right)\left(x+b\right)=x^{2}+\left(a+b\right)x+ab$

$\left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$ $\left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}$

$x^{3}-y^{3}=\left(x-y\right)\left(x^{2}+xy+y^{2}\right)$ $x^{3}+y^{3}=\left(x+y\right)\left(x^{2}-xy+y^{2}\right)$

$a^{3}+b^{3}+c^{3}-3abc=\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)$

Think and Reflect — patterns and signs (pages 69–71)

Sample four-square pattern: $n^{2}-3\left(n+1\right)^{2}+3\left(n+2\right)^{2}-\left(n+3\right)^{2}=0\text{.}$ Expanding gives zero coefficients for $n^{2}$ , n and the constant term. For 1, 4, 9, 16: $1-12+27-16=0$ .

The difference $\left(a+b\right)^{2}-\left(a^{2}+b^{2}\right)$ is 2ab. It is negative exactly when a and b have opposite signs; positive exactly when both are non-zero and have the same sign; zero exactly when at least one is zero. Squares are non-negative, not always positive.

Why

\left(a+b\right)^{2}

contains two ab regions. Area model for positive lengths; symbolic dimensions, not to scale.

Exercise Set 4.1

1. Using the identity $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}\text{,}$ expand the following: (i) $\left(7x+4y\right)^{2}$ (ii) $\left(\left(\frac{7}{5}\right)x+\left(\frac{3}{2}\right)y\right)^{2}$ (iii) $\left(2.5p+1.5q\right)^{2}$ (iv) $\left(\left(\frac{3}{4}\right)s+8t\right)^{2}$ (v) $\left(x+\frac{1}{2y}\right)^{2}$ (vi) $\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$

SOLUTION (i)

49x^{2}+56xy+16y^{2}

. (ii)

\left(\frac{49}{25}\right)x^{2}+\left(\frac{21}{5}\right)xy+\left(\frac{9}{4}\right)y^{2}\text{.}

(iii)

6.25p^{2}+7.5pq+2.25q^{2}

. (iv)

\left(\frac{9}{16}\right)s^{2}+12st+64t^{2}\text{.}

(v)

x^{2}+\frac{x}{y}+\frac{1}{4y^{2}}

. (vi)

\frac{1}{x^{2}}+\frac{2}{xy}+\frac{1}{y^{2}}

Substitution steps: use $A^{2}+2\mathrm{AB}+B^{2}$ with (A,B) respectively (7x,4y), $\left(\frac{7x}{5},\frac{3y}{2}\right)$ , (2.5p,1.5q), $\left(\frac{3s}{4},8t\right)$ , $\left(x,\frac{1}{2y}\right)$ , $\left(\frac{1}{x},\frac{1}{y}\right)$ . The six middle terms are 56xy, $\frac{21xy}{5}$ , 7.5pq, 12st, $\frac{x}{y}$ and $\frac{2}{xy}$ . In (v), $y\ne 0$ ; in (vi), $x\ne 0$ and $y\ne 0$ .

2. Using the same identity, find the values of the following: (i) $\left(64\right)^{2}$ (ii) $\left(105\right)^{2}$ (iii) $\left(205\right)^{2}$

SOLUTION (i)

\begin{aligned}\left(60+4\right)^{2}&=3600+480+16\\&=4096\text{.}\end{aligned}

(ii)

\begin{aligned}\left(100+5\right)^{2}&=10000+1000+25\\&=11025\text{.}\end{aligned}

(iii)

\begin{aligned}\left(200+5\right)^{2}&=40000+2000+25\\&=42025\text{.}\end{aligned}

Think and Reflect What if we replace b by −b in

\left(a+b\right)^{2}=a^{2}+2ab+b^{2}

? Answer. We get

\left(a-b\right)^{2}=a^{2}-2ab+b^{2}

— another identity, used the same way.

Exercise Set 4.2

1. Factor completely: (i) $9x^{2}+24xy+16y^{2}$ (ii) $4s^{2}+20st+25t^{2}$ (iii) $49x^{2}+28xy+4y^{2}$ (iv) $64p^{2}+\left(\frac{32}{3}\right)pq+\left(\frac{4}{9}\right)q^{2}$ *(v) $3a^{2}+4ab+\left(\frac{4}{3}\right)b^{2}$ *(vi) $\left(\frac{9}{5}\right)s^{2}+6sv+5v^{2}$

SOLUTION (i) $\left(3x+4y\right)^{2}$ . (ii) $\left(2s+5t\right)^{2}$ . (iii) $\left(7x+2y\right)^{2}$ . (iv) $\left(8p+\left(\frac{2}{3}\right)q\right)^{2}$

[=\left(\frac{4}{9}\right)\left(12p+q\right)^{2}]

. *(v) Take out

\frac{1}{3}

\begin{aligned}3a^{2}+4ab+\left(\frac{4}{3}\right)b^{2}&=\left(\frac{1}{3}\right)\left(9a^{2}+12ab+4b^{2}\right)\\&=\left(\frac{1}{3}\right)\left(3a+2b\right)^{2}\text{.}\end{aligned}

*(vi) Take out

\frac{1}{5}

\begin{aligned}\left(\frac{9}{5}\right)s^{2}+6sv+5v^{2}&=\left(\frac{1}{5}\right)\left(9s^{2}+30sv+25v^{2}\right)\\&=\left(\frac{1}{5}\right)\left(3s+5v\right)^{2}\text{.}\end{aligned}

For (i)–(iv), identify (A,B) as (3x,4y), (2s,5t), (7x,2y), $\left(8p,\frac{2q}{3}\right)$ . In each case the middle term equals 2AB, so $A^{2}+2\mathrm{AB}+B^{2}=\left(A+B\right)^{2}\text{.}$ In (v) and (vi), first remove the displayed common fractional factor and then apply the same identity.

2. Find the values using the identity $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ : (i) $\left(79\right)^{2}$ (ii) $\left(193\right)^{2}$ (iii) $\left(299\right)^{2}$

SOLUTION (i)

\begin{aligned}\left(80-1\right)^{2}&=6400-160+1\\&=6241\text{.}\end{aligned}

(ii)

\begin{aligned}\left(200-7\right)^{2}&=40000-2800+49\\&=37249\text{.}\end{aligned}

(iii)

\begin{aligned}\left(300-1\right)^{2}&=90000-600+1\\&=89401\text{.}\end{aligned}

Labels for Fig. 4.4: three squares and six rectangles. Area model for positive lengths; symbolic dimensions, not to scale.

The two ab regions, two bc regions and two ac regions give the three doubled cross-terms. The area interpretation uses positive lengths; distributivity proves the identity for all real a, b and c.

Exercise Set 4.3

1. Find the following squares using one of the identities (state which is easier): (i) $117^{2}$ (ii) $78^{2}$ (iii) $198^{2}$ (iv) $214^{2}$ (v) $1104^{2}$ (vi) $1120^{2}$

SOLUTION (i)

\begin{aligned}\left(120-3\right)^{2}&=14400-720+9\\&=13689\text{.}\end{aligned}

(ii)

\begin{aligned}\left(80-2\right)^{2}&=6400-320+4\\&=6084\text{.}\end{aligned}

(iii)

\begin{aligned}\left(200-2\right)^{2}&=40000-800+4\\&=39204\text{.}\end{aligned}

(iv)

\begin{aligned}\left(200+14\right)^{2}&=40000+5600+196\\&=45796\text{.}\end{aligned}

(v)

\begin{aligned}\left(1100+4\right)^{2}&=1210000+8800+16\\&=1218816\text{.}\end{aligned}

(vi)

\begin{aligned}\left(1100+20\right)^{2}&=1210000+44000+400\\&=1254400\text{.}\end{aligned}

2. Factor using suitable identities: (i) $16y^{2}-24y+9$ (ii) $\left(\frac{9}{4}\right)s^{2}+6st+4t^{2}$ (iii) $\frac{m^{2}}{9}+\frac{mk}{3}+\frac{k^{2}}{4}+3nk+2mn+9n^{2}$ (iv) $\frac{p^{2}}{16}-2+\frac{16}{p^{2}}$ (v) $9a^{2}+4b^{2}+c^{2}-12ab+6ac-4bc$

SOLUTION (i) $\left(4y-3\right)^{2}$ . (ii) $\left(\left(\frac{3}{2}\right)s+2t\right)^{2}$ . (iii) $\left(\frac{m}{3}+\frac{k}{2}+3n\right)^{2}$ . (iv) $\left(\frac{p}{4}-\frac{4}{p}\right)^{2}$ . (v) $\left(3a-2b+c\right)^{2}$ .

Matching bases: (i) $A=4y$ , $B=3$ , giving $\left(A-B\right)^{2}$ ; (ii) $A=\frac{3s}{2}$ , $B=2t$ , giving $\left(A+B\right)^{2}$ ; (iii) $A=\frac{m}{3}$ , $B=\frac{k}{2}$ , $C=3n$ , whose doubled pair-products are $\frac{mk}{3}$ , 2mn and 3nk; (iv) $A=\frac{p}{4}$ , $B=\frac{4}{p}$ , giving $2\mathrm{AB}=2$ and requiring $p\ne 0$ ; (v) $A=3a$ , $B=-2b$ , $C=c$ .

3. Expand using $\left(a+b+c\right)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$ : (i) $\left(p+3q+7r\right)^{2}$ (ii) $\left(3x-2y+4z\right)^{2}$

SOLUTION (i)

p^{2}+9q^{2}+49r^{2}+6pq+42qr+14pr\text{.}

(ii)

9x^{2}+4y^{2}+16z^{2}-12xy-16yz+24xz\text{.}

(i) The doubled products are $2\left(p\right)\left(3q\right)=6pq$ , $2\left(3q\right)\left(7r\right)=42qr$ and $2\left(7r\right)\left(p\right)=14pr$ . (ii) They are $2\left(3x\right)\left(-2y\right)=-12xy$ , $2\left(-2y\right)\left(4z\right)=-16yz$ and $2\left(4z\right)\left(3x\right)=24xz$ . Add these to the three squared terms.

4. Is this an identity? $\left(a+b-c\right)^{2}+\left(a-b+c\right)^{2}+\left(a-b-c\right)^{2}=2a^{2}+2b^{2}+2c^{2}\text{.}$

SOLUTION Expanding the left side gives

3a^{2}+3b^{2}+3c^{2}-2ab-2bc-2ca\text{,}

which is not equal to

2a^{2}+2b^{2}+2c^{2}

. So No, it is not an identity (it holds only for special values, not all a, b, c).

A single counterexample also settles it: $a=1$ , $b=c=0$ gives $\mathrm{LHS}=3$ but $\mathrm{RHS}=2$ . An identity must hold for every permitted choice.

Think and Reflect Evaluate

35^{2}

65^{2}

85^{2}

105^{2}

. Do you observe a pattern? Answer.

35^{2}=1225

65^{2}=4225

85^{2}=7225

105^{2}=11025

. For a number ending in 5, say “n5”, the square is n(n + 1) followed by 25 — e.g.

85^{2}

8\times 9=72

, then 25 → 7225.

Think and Reflect — Fig. 4.6 and algebra tiles (pages 78–80)

Fig. 4.6 represents $\left(a+b+c\right)^{2}+\left(a+b-c\right)^{2}+\left(a-b+c\right)^{2}+\left(a-b-c\right)^{2}=4a^{2}+4b^{2}+4c^{2}\text{.}$ Expanding cancels all cross-terms. The area arrangement assumes positive indicated side lengths.

Splitting 7x into 2x+5x leaves only $2\times 5=10$ unit spaces, not the required 12. Splitting it into 3x+4x gives both $3+4=7$ and $3\times 4=12$ . Both conditions must hold.

$\left(x+2\right)\left(x+3\right)=x^{2}+5x+6$ . To arrange $x^{2}+11x+30$ , put five x-strips along one side and six along the other, with a $5\times 6$ block of unit tiles. The factors are (x+5)(x+6).

Generally $\left(x+a\right)\left(x+b\right)=x^{2}+\left(a+b\right)x+ab$ . Also $\left(2x+3\right)\left(3x+1\right)=6x^{2}+11x+3$ . The blanks in (px+a)(qx+b) are pq, pb+aq, ab, because distributing gives pqx²+pbx+aqx+ab.

Exercise Set 4.4

1. Fill in the blanks to complete the following identities: (i) s² − 11s + 24 = (____)(____) (ii) (____)(x + 1) = (3x² − 4x − 7) (iii) 10x² − 11x − 6 = (2x − ___)(___ + 2) (iv) 6x² + 7x + 2 = (____)(____)

SOLUTION (i) (s − 3)(s − 8). (ii) (3x − 7)(x + 1). (iii) (2x − 3)(5x + 2). (iv) (2x + 1)(3x + 2).

The middle-term splits are: (i) $-11s=-3s-8s$ ; (ii) $-4x=3x-7x$ ; (iii) $-11x=4x-15x$ ; (iv) $7x=4x+3x$ . Grouping, for example, $\begin{aligned}10x^{2}+4x-15x-6&=2x\left(5x+2\right)-3\left(5x+2\right)\\&=\left(2x-3\right)\left(5x+2\right)\text{.}\end{aligned}$ The blanks in (iii) are 3 and 5x.

2. Use a suitable identity to find these products without multiplying directly: (i) $\left(41\right)^{2}$ (ii) $\left(27\right)^{2}$ (iii) $23\times 17$ (iv) $\left(135\right)^{2}$ (v) $\left(97\right)^{2}$ (vi) $18\times 29$ (vii) $34\times 43$ (viii) $\left(205\right)^{2}$

SOLUTION (i)

\left(40+1\right)^{2}=1681

. (ii)

\left(30-3\right)^{2}=729

. (iii)

\begin{aligned}\left(20+3\right)\left(20-3\right)&=400-9\\&=391\text{.}\end{aligned}

(iv)

\left(130+5\right)^{2}=18225

. (v)

\left(100-3\right)^{2}=9409

. (vi)

\begin{aligned}18\times 29&=\left(23.5\right)^{2}-\left(5.5\right)^{2}\\&=552.25-30.25\\&=522\text{.}\end{aligned}

(vii)

\begin{aligned}34\times 43&=\left(38.5\right)^{2}-\left(4.5\right)^{2}\\&=1462\text{.}\end{aligned}

(viii)

\left(200+5\right)^{2}=42025

An integer-only alternative for (vi) is $\begin{aligned}\left(20-2\right)\left(20+9\right)&=20^{2}+\left(-2+9\right)20-18\\&=400+140-18\\&=522\text{.}\end{aligned}$ For (vii), $\begin{aligned}\left(40-6\right)\left(40+3\right)&=1600-120-18\\&=1462\text{.}\end{aligned}$ These use $\left(x+a\right)\left(x+b\right)=x^{2}+\left(a+b\right)x+ab$ .

3. Factor the following: (i) $9a^{2}+b^{2}+4c^{2}-6ab+12ac-4bc$ (ii) $16s^{2}+25t^{2}-40st$ (iii) $r^{2}-r-42$ (iv) $49g^{2}+14gh+h^{2}$ (v) $64u^{2}+121v^{2}+4w^{2}-176uv-32uw+44vw$

SOLUTION (i) $\left(3a-b+2c\right)^{2}$ . (ii) $\left(4s-5t\right)^{2}$ . (iii) (r − 7)(r + 6). (iv) $\left(7g+h\right)^{2}$ . (v) $\left(8u-11v-2w\right)^{2}$ .

For (i), square the three terms 3a, −b and 2c and check their doubled products −6ab, 12ac and −4bc. For (ii) use $\left(4s-5t\right)^{2}$ . For (iii), $-7+6=-1$ and $\left(-7\right)6=-42$ . For (iv) use $\left(7g+h\right)^{2}$ . For (v), the bases 8u, −11v, −2w give doubled products −176uv, −32uw and +44vw.

Think and Reflect James writes

\left(a-b\right)^{2}\left(a+b\right)=\left(a^{2}-2ab+b^{2}\right)\left(a+b\right)\text{;}

Reshma writes it as

\left(a-b\right)[\left(a-b\right)\left(a+b\right)]=\left(a-b\right)\left(a^{2}-b^{2}\right)\text{.}

Who is correct? Answer. Both are correct — they are two valid routes to the same product; Reshma’s grouping is just a quicker way, and both give

a^{3}-a^{2}b-ab^{2}+b^{3}

Think and Reflect — further identities and cancellation

$x^{4}-y^{4}=\left(x-y\right)\left(x+y\right)\left(x^{2}+y^{2}\right)\text{,}$ and $x^{5}-y^{5}=\left(x-y\right)\left(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4}\right)\text{.}$ Expanding makes the intermediate terms cancel. Thus x−y is a factor in both cases.

For the page 87 fraction, $t^{2}+2ts-48s^{2}=\left(t+8s\right)\left(t-6s\right)\text{.}$ The numerator $\left(6s-t\right)^{2}=\left(t-6s\right)^{2}$ . Cancelling gives $\frac{t-6s}{t+8s}$ , with the original restrictions $t\ne 6s$ and $t\ne -8s$ .

Exercise Set 4.5

1. Simplify the following rational expressions (denominators ≠ 0): (i) $\frac{3p^{2}-3pq-18q^{2}}{p^{2}+3pq-10q^{2}}$ (ii) $\frac{n^{3}-3n^{2}m+3nm^{2}-m^{3}}{5m^{2}-10mn+5n^{2}}$ (iii) $\frac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}$ (iv) $\frac{4y^{2}-20yz+25z^{2}}{25z^{2}-4y^{2}}$ (v) $\frac{\left(x^{2}+x-6\right)\left(x^{2}-7x+12\right)}{\left(x^{2}-6x+8\right)\left(x^{2}-9\right)}$ (vi) $\frac{p^{4}-16}{p^{2}-4p+4}$

SOLUTION (i) Factor: numerator = 3(p − 3q)(p + 2q), denominator = (p + 5q)(p − 2q). They share no common factor, so the expression is already in lowest terms: $\frac{3\left(p-3q\right)\left(p+2q\right)}{[\left(p+5q\right)\left(p-2q\right)]}$ . (ii) Numerator =

\left(n-m\right)^{3}

, denominator =

5\left(n-m\right)^{2}

⇒ $\frac{n-m}{5}$ . (iii) Numerator =

\left(w-v+x\right)\left(w^{2}+v^{2}+x^{2}+wv+vx-wx\right)\text{,}

denominator =

\left(w-v+x\right)^{2}

⇒ $\frac{w^{2}+v^{2}+x^{2}+wv+vx-wx}{w-v+x}$ . (iv) Numerator =

\left(5z-2y\right)^{2}

, denominator = (5z − 2y)(5z + 2y) ⇒ $\frac{5z-2y}{5z+2y}$ . (v) =

\frac{[\left(x+3\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)]}{[\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)]}=1\text{.}

(vi)

p^{4}-16=\left(p-2\right)\left(p+2\right)\left(p^{2}+4\right)\text{,}

denominator =

\left(p-2\right)^{2}

⇒ $\frac{\left(p+2\right)\left(p^{2}+4\right)}{p-2}$ .

Keep all original exclusions: (i) $p\ne -5q$ and $p\ne 2q$ ; (ii) $n\ne m$ ; (iii) $w-v+x\ne 0$ ; (iv) $5z-2y\ne 0$ and $5z+2y\ne 0$ ; (v) $x\ne 2$ , 4, 3, −3; (vi) $p\ne 2$ . A cancellation never restores an excluded input.

In-text construction — Saira’s rectangle

One

x^{2}

tile, eight x tiles and fifteen unit tiles. Symbolic area model, not to scale.

$x^{2}+8x+15=\left(x+3\right)\left(x+5\right)$ . Arrange the $x^{2}$ square, three x-strips along one side, five along the other and a $3\times 5$ block of unit squares. This gives dimensions x+3 and x+5 for $x>0$ .

End-of-Chapter Exercises — Questions 1–13

1. Use suitable identities to find the following products: (i) $\left(-3x+4\right)^{2}$ (ii) (2s + 7)(2s − 7) (iii) $\left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)$ (iv) (2n + 7)(2n − 7) (v) $\left(s-2t\right)\left(s^{2}+2st+4t^{2}\right)$ (vi) $\left(\frac{1}{2r}-4r\right)^{2}$ (vii) $\left(-3m+4k-l\right)^{2}$ (viii) $\left(x-\left(\frac{1}{3}\right)y\right)^{3}$ (ix) $\left(\left(\frac{7}{2}\right)k-\left(\frac{2}{3}\right)m\right)^{3}$

SOLUTION (i)

9x^{2}-24x+16

. (ii)

4s^{2}-49

. (iii)

p^{4}-\frac{1}{4}

. (iv)

4n^{2}-49

. (v)

s^{3}-8t^{3}

[(a − b)(a² + ab + b²) with

a=s

, b = 2t]. (vi)

16r^{2}-4+\frac{1}{4r^{2}}

. (vii)

9m^{2}+16k^{2}+l^{2}-24mk-8kl+6lm\text{.}

(viii)

x^{3}-x^{2}y+\left(\frac{1}{3}\right)xy^{2}-\left(\frac{1}{27}\right)y^{3}\text{.}

(ix)

\left(\frac{343}{8}\right)k^{3}-\left(\frac{49}{2}\right)k^{2}m+\left(\frac{14}{3}\right)km^{2}-\left(\frac{8}{27}\right)m^{3}\text{.}

For (i) use $A=-3x$ , $B=4$ in $\left(A+B\right)^{2}$ . Parts (ii)–(iv) use $A^{2}-B^{2}$ . In (vi), $2\cdot [\frac{1}{2r}]\cdot 4r=4$ , with $r\ne 0$ . For (vii), take $A=-3m$ , $B=4k$ , $C=-l$ . In (viii), the cubic middle terms are $-3x^{2}\left(\frac{y}{3}\right)=-x^{2}y$ and $3x\left(\frac{y}{3}\right)^{2}=\frac{xy^{2}}{3}$ . In (ix), they are $-3\left(\frac{7k}{2}\right)^{2}\left(\frac{2m}{3}\right)=-\frac{49k^{2}m}{2}$ and $3\left(\frac{7k}{2}\right)\left(\frac{2m}{3}\right)^{2}=\frac{14km^{2}}{3}\text{.}$

2. Find the values using suitable identities: (i) $17\times 21$ (ii) $104\times 96$ (iii) $24\times 16$ (iv) $147^{3}$ (v) $199^{3}$ (vi) $127^{3}$ (vii) $\left(-107\right)^{3}$ (viii) $\left(-299\right)^{3}$

SOLUTION (i)

\begin{aligned}\left(19-2\right)\left(19+2\right)&=361-4\\&=357\text{.}\end{aligned}

(ii)

\left(100+4\right)\left(100-4\right)=9984

. (iii)

\left(20+4\right)\left(20-4\right)=384

. (iv)

\left(150-3\right)^{3}=3176523

. (v)

\left(200-1\right)^{3}=7880599

. (vi)

\left(130-3\right)^{3}=2048383

. (vii)

-\left(107^{3}\right)=-1225043

. (viii)

-\left(299^{3}\right)=-26730899

Cubic working: $\begin{aligned}147^{3}&=150^{3}-3\cdot 150^{2}\cdot 3+3\cdot 150\cdot 3^{2}-3^{3}\\&=3{,}375{,}000-202{,}500+4{,}050-27\\&=3{,}176{,}523\text{.}\end{aligned}$ Similarly $\begin{aligned}199^{3}&=8{,}000{,}000-120{,}000+600-1\\&=7{,}880{,}599\text{;}\end{aligned}$ $\begin{aligned}127^{3}&=2{,}197{,}000-152{,}100+3{,}510-27\\&=2{,}048{,}383\text{.}\end{aligned}$

$\begin{aligned}107^{3}&=1{,}000{,}000+210{,}000+14{,}700+343\\&=1{,}225{,}043\text{,}\end{aligned}$ so $\left(-107\right)^{3}=-1{,}225{,}043$ . Also $\begin{aligned}299^{3}&=27{,}000{,}000-270{,}000+900-1\\&=26{,}730{,}899\text{,}\end{aligned}$ so $\left(-299\right)^{3}=-26{,}730{,}899$ .

3. Factor the following algebraic expressions: (i) $4y^{2}+1+\frac{1}{16y^{2}}$ (ii) $9m^{2}-\frac{1}{25n^{2}}$ (iii) $27b^{3}-\frac{1}{64b^{3}}$ (iv) $x^{2}+\frac{5x}{6}+\frac{1}{6}$ (v) $27u^{3}-\frac{1}{125}-\frac{27u^{2}}{5}+\frac{9u}{25}$ (vi) $64y^{3}+\left(\frac{1}{125}\right)z^{3}$ (vii) $p^{3}+27q^{3}+r^{3}-9$ pqr (viii) $9m^{2}-12m+4$ (ix) $9x^{3}-\left(\frac{8}{3}\right)y^{3}+\left(\frac{1}{3}\right)z^{3}+6$ xyz (x) $4x^{2}+9y^{2}+36z^{2}+12xz+36yz+24xy$ (xi) $27u^{3}-\frac{1}{216}-\frac{9u^{2}}{2}+\frac{u}{4}$

SOLUTION (i) $\left(2y+\frac{1}{4y}\right)^{2}$ . (ii) $\left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)$ . (iii) $\left(3b-\frac{1}{4b}\right)\left(9b^{2}+\frac{3}{4}+\frac{1}{16b^{2}}\right)$ . (iv) $\left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right)$ . (v) $\left(3u-\frac{1}{5}\right)^{3}$ . (vi) $\left(4y+\frac{z}{5}\right)\left(16y^{2}-\frac{4yz}{5}+\frac{z^{2}}{25}\right)$ . (vii) $\left(p+3q+r\right)\left(p^{2}+9q^{2}+r^{2}-3pq-3qr-pr\right)$ . (viii) $\left(3m-2\right)^{2}$ . (ix) Multiply through by 3:

27x^{3}-8y^{3}+z^{3}+18

xyz =

\left(3x-2y+z\right)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right)\text{.}

So the expression =

\left(\frac{1}{3}\right)\left(3x-2y+z\right)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right)\text{.}

(x) Textbook inconsistency: the printed expression has cross-terms 12xz + 36yz + 24xy. It is not

\left(2x+3y+6z\right)^{2}

, whose cross-terms are 12xy + 24xz + 36yz. For

x=y=1

z=0

, the printed expression is 37, while that proposed square is 25. A possible corrected question would swap the coefficients of xy and xz; this is a proposed correction, not the answer to the expression as printed. (xi) $\left(3u-\frac{1}{6}\right)^{3}$ .

Checks and restrictions: (i) $2\cdot \left(2y\right)\cdot [\frac{1}{4y}]=1$ , $y\ne 0$ ; (ii) use a difference of squares, $n\ne 0$ ; (iii) use a difference of cubes with $A=3b$ and $B=\frac{1}{4b}$ , $b\ne 0$ . In (iv), $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ and $\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{6}$ . Parts (v) and (xi) are $\left(A-B\right)^{3}$ with $\left(A,B\right)=\left(3u,\frac{1}{5}\right)$ and $\left(3u,\frac{1}{6}\right)$ . For (vii) use $A=p$ , $B=3q$ , $C=r$ in the three-cube identity; for (ix) first multiply by 3 and use $A=3x$ , $B=-2y$ , $C=z$ .

4. Simplify the following (denominators ≠ 0): (i) $\frac{4x^{2}+4x+1}{4x^{2}-1}$ (ii) $\frac{9\left(3a^{3}-24b^{3}\right)}{9a^{2}-36b^{2}}$ (iii) $\frac{s^{3}+125t^{3}}{s^{2}-2st-35t^{2}}$

SOLUTION (i)

\frac{\left(2x+1\right)^{2}}{[\left(2x-1\right)\left(2x+1\right)]}=\frac{2x+1}{2x-1}\text{.}

(ii)

\frac{27\left(a-2b\right)\left(a^{2}+2ab+4b^{2}\right)}{[9\left(a-2b\right)\left(a+2b\right)]}=\frac{3\left(a^{2}+2ab+4b^{2}\right)}{a+2b}\text{.}

(iii)

\frac{\left(s+5t\right)\left(s^{2}-5st+25t^{2}\right)}{[\left(s-7t\right)\left(s+5t\right)]}=\frac{s^{2}-5st+25t^{2}}{s-7t}\text{.}

Original restrictions: (i) $x\ne \frac{1}{2}$ and $x\ne -\frac{1}{2}$ ; (ii) $a\ne 2b$ and $a\ne -2b$ ; (iii) $s\ne 7t$ and $s\ne -5t$ . These remain necessary even when the corresponding factor cancels.

5. Find possible expressions for the length and breadth of rectangles with these areas: (i) $25a^{2}-30ab+9b^{2}$ (ii) $36s^{2}-49t^{2}$

SOLUTION (i)

25a^{2}-30ab+9b^{2}=\left(5a-3b\right)^{2}

⇒ length = breadth = (5a − 3b). (ii)

36s^{2}-49t^{2}=\left(6s+7t\right)\left(6s-7t\right)

⇒ length = (6s + 7t), breadth = (6s − 7t).

Dimensions must be positive. In (i), use 5a−3b only when it is positive; more generally the square has side |5a−3b|, provided it is non-zero. The pair in (ii) works when both 6s+7t and 6s−7t are positive. Each dimension is measured in units, and the product in square units.

6. Find possible expressions for the length, breadth and height of cuboids with these volumes: (i) $6a^{2}-24b^{2}$ (ii) $3ps^{2}-15ps+12p$

SOLUTION (i)

6a^{2}-24b^{2}=6\left(a-2b\right)\left(a+2b\right)

⇒ dimensions 6, (a − 2b), (a + 2b). (ii)

3ps^{2}-15ps+12p=3p\left(s-1\right)\left(s-4\right)

⇒ dimensions 3p, (s − 1), (s − 4).

These are sample factor-based dimensions in units. They describe physical cuboids only when all chosen factors are positive: for example $a>2|b|$ for (i), and $p>0$ , $s>4$ for (ii). Other positive factor groupings can also work.

7. The village playground is a square of side 40 m. A path of width s metres is created around it. Find an expression for the area of the path in terms of s.

SOLUTION Outer square side = (40 + 2s). Path area =

\begin{aligned}\left(40+2s\right)^{2}-40^{2}&=1600+160s+4s^{2}-1600\\&=4s^{2}+160s\end{aligned}

\left(=4s\left(s+40\right)\right)sq

. m.

Placement assumption: the path is outside the existing 40 m square and has constant width $s\ge 0$ . If a path of width s were instead inside the boundary, its area would be $40^{2}-\left(40-2s\right)^{2}=160s-4s^{2}$ for $0\le s\le 20$ . The two interpretations should not be mixed.

8. If a number plus its reciprocal equals $\frac{10}{3}$ , find the number.

SOLUTION

\begin{aligned}x+\frac{1}{x}&=\frac{10}{3}\\\Rightarrow 3x^{2}-10x+3&=0\\\Rightarrow \left(3x-1\right)\left(x-3\right)&=0\\\Rightarrow x&=3\end{aligned}

\frac{1}{3}

9. A rectangular pool has area $2x^{2}+7x+3$ square hastas. If its width is 2x + 1 hastas, find its length.

SOLUTION

2x^{2}+7x+3=\left(2x+1\right)\left(x+3\right)\text{.}

Dividing by the width (2x + 1) gives length = (x + 3) hastas.

For a physical pool, width $2x+1>0$ and length $x+3>0$ ; together these require $x>-\frac{1}{2}$ .

*10. If both x − 2 and $x-\frac{1}{2}$ are factors of $px^{2}+5x+r$ , show that $p=r$ .

SOLUTION

x=2

is a root:

\begin{aligned}4p+10+r&=0\\\Rightarrow 4p+r&=-10\text{.}\end{aligned}

x=\frac{1}{2}

is a root:

\begin{aligned}\frac{p}{4}+\frac{5}{2}+r&=0\\\Rightarrow p+4r&=-10\text{.}\end{aligned}

Subtracting the two equations:

3p-3r=0

⇒ $p=r$ .

Substituting $p=r$ into $4p+r=-10$ also gives $p=r=-2$ . Check: $-2x^{2}+5x-2=-2\left(x-2\right)\left(x-\frac{1}{2}\right)$ .

*11. If $a+b+c=5$ and $ab+bc+ca=10$ , prove that $a^{3}+b^{3}+c^{3}-3$ abc = −25.

SOLUTION

a^{3}+b^{3}+c^{3}-3

abc =

\begin{aligned}\left(a+b+c\right)[\left(a+b+c\right)^{2}-3\left(ab+bc+ca\right)]&=5[25-30]\\&=5\left(-5\right)\\&=-25\text{.}\end{aligned}

✓

Consistency note: the algebraic implication above is valid, but the given conditions cannot hold for real a,b,c. They give $a^{2}+b^{2}+c^{2}=25-20=5$ , whereas $\begin{aligned}\left(a-b\right)^{2}+\left(b-c\right)^{2}+\left(c-a\right)^{2}&=2\left(5-10\right)\\&=-10\text{,}\end{aligned}$ impossible for real squares. This is an issue with the question’s premises, not a real numerical example.

*12. By factoring, check that $n^{3}-n$ is always divisible by 6 for all natural numbers n.

SOLUTION

\begin{aligned}n^{3}-n&=n\left(n^{2}-1\right)\\&=\left(n-1\right)\cdot n\cdot \left(n+1\right)\end{aligned}

— the product of three consecutive integers. Among any three consecutive integers, one is divisible by 3 and at least one is even (divisible by 2). So the product is divisible by

2\times 3=6

*13. Find the value of: (i) $x^{3}+y^{3}-12xy+64$ , when $x+y=-4$ (ii) $x^{3}-8y^{3}-36xy-216$ , when $x=2y+6$

SOLUTION (i)

x^{3}+y^{3}+4^{3}-3\cdot x\cdot y\cdot 4=\left(x+y+4\right)\left(\ldots \right)\text{.}

Since

x+y=-4

x+y+4=0

, so the value is 0. (ii)

x^{3}+\left(-2y\right)^{3}+\left(-6\right)^{3}-3\cdot x\cdot \left(-2y\right)\cdot \left(-6\right)=\left(x-2y-6\right)\left(\ldots \right)\text{.}

Since

x=2y+6

x-2y-6=0

, so the value is 0.

Common Mistakes to Avoid

Watch out for these

Writing $\left(a+b\right)^{2}=a^{2}+b^{2}$ — the middle term 2ab must be there.
Sign of the middle term: $\left(a-b\right)^{2}=a^{2}$ − $2ab+b^{2}$ .
In $\left(a+b+c\right)^{2}$ , forgetting the three cross terms 2ab, 2bc, 2ca (mind their signs).
For $\left(a\pm b\right)^{3}$ , missing the $3a^{2}b$ and $3ab^{2}$ terms (coefficients 1, 3, 3, 1).
When splitting the middle term, both sum and product conditions must hold together.
While simplifying rational expressions, cancel only common factors, never individual terms.

Practice MCQs & Assertion–Reason

Additional ClearStudy practice, separate from the NCERT exercises.

1. $\left(a+b\right)^{2}$ equals:

(a) $a^{2}+b^{2}$ (b) $a^{2}+2ab+b^{2}$ (c) $a^{2}-2ab+b^{2}$ (d) 2ab

2. $\left(a-b\right)^{2}$ equals:

(a) $a^{2}-b^{2}$ (b) $a^{2}+2ab+b^{2}$ (c) $a^{2}-2ab+b^{2}$ (d) $a^{2}-b^{2}+2ab$

3. $a^{2}-b^{2}$ equals:

(a) $\left(a-b\right)^{2}$ (b) $\left(a+b\right)^{2}$ (c) (a + b)(a − b) (d) (a + b)(a + b)

4. (x + 3)(x + 4) equals:

(a) $x^{2}+12$ (b) $x^{2}+7x+12$ (c) $x^{2}+12x+7$ (d) $x^{2}+7x+7$

5. The factors of $x^{2}-9$ are:

(a) (x − 3)(x − 3) (b) (x + 3)(x + 3) (c) (x − 3)(x + 3) (d) (x − 9)(x + 1)

6. $\left(a+b\right)^{3}$ equals:

(a) $a^{3}+b^{3}$ (b) $a^{3}+3a^{2}b+3ab^{2}+b^{3}$ (c) $a^{3}-b^{3}$ (d) $a^{3}+3ab+b^{3}$

7. Using an identity, $99^{2}$ equals:

(a) 9801 (b) 9081 (c) 9999 (d) 9901

8. $x^{2}+10x+25$ factorises to:

(a) $\left(x+5\right)^{2}$ (b) $\left(x-5\right)^{2}$ (c) (x + 25)(x + 1) (d) (x + 5)(x − 5)

9. $\left(x-y\right)\left(x^{2}+xy+y^{2}\right)$ equals:

(a) $x^{3}+y^{3}$ (b) $x^{3}-y^{3}$ (c) $\left(x-y\right)^{3}$ (d) $x^{2}-y^{2}$

10. If $a+b=5$ and $ab=6$ , then $a^{2}+b^{2}$ equals:

(a) 11 (b) 13 (c) 25 (d) 37

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(c), 6-(b), 7-(a), 8-(a), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}\text{.}$

Reason: Distributing (a + b)(a + b) gives $a^{2}+ab+ba+b^{2}=a^{2}+2ab+b^{2}\text{.}$

A-R 2. Assertion: $x^{2}-16=\left(x-4\right)\left(x+4\right)$ .

Reason: $a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\text{.}$

A-R 3. Assertion: $\left(a-b\right)^{2}=a^{2}+2ab+b^{2}\text{.}$

Reason: The square of (a − b) expands to $a^{2}-2ab+b^{2}$ .

A-R 4. Assertion: $x^{2}+5x+6=\left(x+2\right)\left(x+3\right)\text{.}$

Reason: We need two numbers with sum 5 and product 6, namely 2 and 3.

A-R 5. Assertion: If $a+b+c=0$ , then $a^{3}+b^{3}+c^{3}=3$ abc.

Reason: $a^{3}+b^{3}+c^{3}-3$ abc = $\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)\text{.}$

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

An identity is true for all values of the variables.
Squares: $\left(a\pm b\right)^{2}=a^{2}\pm 2ab+b^{2}\text{;}$ $\left(a+b+c\right)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\text{.}$
Difference of squares: $a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\text{.}$
Product form: $\left(x+a\right)\left(x+b\right)=x^{2}+\left(a+b\right)x+ab$ — basis of “splitting the middle term”.
Cubes: $\left(a\pm b\right)^{3}$ ; x³ ± y³ = (x ± y)(x² ∓ $xy+y^{2}$ ).
$a^{3}+b^{3}+c^{3}-3$ abc = $\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)\text{;}$ if $a+b+c=0$ then $a^{3}+b^{3}+c^{3}=3$ abc.
Simplify rational expressions by factorising and cancelling common factors.

Presenting clear mathematical working

Memorise the identities and always write the one you are using before substituting. For calculations like $117^{2}$ , split the number as (round ± small). For factorising, first check for a common factor and a recognisable identity; otherwise split the middle term. For rational expressions, factor numerator and denominator fully before cancelling.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 4 about?

It develops the algebraic identities — squares, difference of squares, and cubes, including $a^{3}+b^{3}+c^{3}-3$ abc — and uses them to expand, factorise and simplify expressions.

What is the identity for (a + b) squared?

$\left(a+b\right)^{2}=a^{2}+2ab+b^{2}\text{.}$

What is

a^{3}+b^{3}+c^{3}-3

abc?

It equals $\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)\text{.}$ If $a+b+c=0$ , then $a^{3}+b^{3}+c^{3}=3$ abc.

Are these Class 9 Maths Ganita Manjari Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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