Exploring Algebraic Identities Class 9 Maths Ganita Manjari Part 1 Chapter 4 NCERT Solutions

Exploring Algebraic Identities Class 9 Maths Ganita Manjari Part 1 Chapter 4 NCERT Solutions Looking for the Exploring Algebraic Identities Class 9 Maths Ganita Manjari Part 1 Chapter 4 NCERT Solutions? You are in the right place. This chapter helps students understand important algebraic identities, their derivations, and their applications in solving mathematical problems quickly and accurately.

Exercise Set 4.1

1. Using the identity (a + b)² = a² + 2ab + b², expand the following:
(i) (7x + 4y)²
(ii) ${(\frac{7}{5}x+\frac{3}{2}y)}^2$
(iii) (2.5p + 1.5q)²
(iv) ${(\frac{3}{4}s+8t)}^2$
(v) ${(x+\frac{1}{2y})}^2$
(vi) ${(\frac{1}{x}+\frac{1}{y})}^2$
Solution:
(i) (7x + 4y)²
Using (a + b)² = a² + 2ab + b²
= (7x)² + 2(7x)(4y) + (4y)²
= 49x² + 56x + 16y²

(ii) ${(\frac{7}{5}x+\frac{3}{2}y)}^2$
Using (a + b)² = a² + 2ab + b²
= ${\left(\frac{7}{5}x\right)}^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+{\left(\frac{3}{2}y\right)}^2$
= $\frac{49}{25}{x}^2+\frac{42}{10}xy+\frac{9}{4}{y}^2$
= $\frac{49}{25}{x}^2+\frac{21}{5}xy+\frac{9}{4}{y}^2$

(iii) (2.5p + 1.5q)²
Using (a + b)² = a² + 2ab + b²
= (2.5p)² + 2 (2.5p)(1.5q) + (1.5q)²
= 6.25p² + 7.5pq + 2.25q²

(iv) (34s+8t)2
Using (a + b)² = a² + 2ab + b²
= ${\left(\frac{3}{4}s\right)}^2+2\left(\frac{3}{4}s\right)\left(8t\right)+{\left(8t\right)}^2$
= $\frac{9}{16}{s}^2+\frac{48}{4}st+64t^2$
= $\frac{9}{16}{s}^2+12st+64t^2$

(v) ${(x+\frac{1}{2y})}^2$
Using (a + b)² = a² + 2ab + b²
= ${\left(x\right)}^2+2\left(x\right)\left(\frac{1}{2y}\right)+{\left(\frac{1}{2y}\right)}^2$
= $x^2+\frac{x}{y}+\frac{1}{4y^2}$

(vi) ${(\frac{1}{x}+\frac{1}{y})}^2$
Using (a + b)² = a² + 2ab + b²
= ${\left(\frac{1}{x}\right)}^2+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+{\left(\frac{1}{y}\right)}^2$
= $\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}$

2. Using the same identity, find the values of the following:
(i) (64)² (ii) (105)² (iii) (205)²
Solution:
(i) (64)²
= (60 + 4)²
Using (a + b)² = a² + 2ab + b²
= (60)² + 2(60)(4) + (4)²
= 3600 + 480 + 16
= 4096

(ii) (105)²
= (100 + 5)²
Using (a + b)² = a² + 2ab + b²
= (100)² + 2(100)(5) + (5)²
= 10000 + 1000 + 25
= 11025

(iii) (205)²
= (200 + 5)²
Using the identity (a + b)² = a² + 2ab + b²
= (200)² + 2(200)(5) + (5)²
= 40000 + 2000 + 25
= 42025

Exercise Set 4.2

1. Factor completely:
(i)  9x² + 24xy + 16y²
(ii)  4s² + 20st + 25t²
(iii) 49x² + 28xy + 4y²
(iv) 64p² + $\frac{32}{3}$pq + $\frac{4}{9}$q²
(v) 3a² + 4ab + $\frac{4}{3}$b²
(vi) $\frac{9}{5}$s² + 6sv + 5v²
(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)
Solution:
(i)  9x² + 24xy + 16y²
= (3x)² + 2(3x)(4y) + (4y)²
Using a² + 2ab + b² = (a + b)²
= (3x + 4y)²
Therefore, 9x² + 24xy + 16y² = (3x + 4y)²

(ii)  4s² + 20st + 25t²
= (2s)² + 2(2s)(5t) + (5t)²
Using a² + 2ab + b² = (a + b)²
= (2s + 5t)²
Therefore, 4s² + 20st + 25t² = (2s + 5t)²

(iii) 49x² + 28xy + 4y²
= (7x)² + 2(7x)(2y) + (2y)²
Using a² + 2ab + b² = (a + b)²
= (7x + 2y)²
Therefore, 49x² + 28xy + 4y² = (7x + 2y)²

(iv) 64p² + $\frac{32}{3}$pq + $\frac{4}{9}$q²
= (8p)² + 2(8p)$\left(\frac{2}{3}q\right)$ + ${\left(\frac{2}{3}q\right)}^2$
Using a² + 2ab + b² = (a + b)²
= (8p + $\frac{2}{3}$q)²
Therefore, 64p² + $\frac{32}{3}$pq + $\frac{4}{9}$q² = (8p + $\frac{2}{3}$q)²

(v) 3a² + 4ab + $\frac{4}{3}$b²
Taking out 3 as a factor,
= $3(a^2+\frac{4}{3}ab+\frac{4}{9}{b}^2)$
= $3[(a{)}^2+2(a)\left(\frac{2}{3}b\right)+{\left(\frac{2}{3}b\right)}^2]$
Using a² + 2ab + b² = (a + b)²
= $3{(a+\frac{2}{3}b)}^2$
Therefore, 3a² + 4ab + $\frac{4}{3}$b² = $3{(a+\frac{2}{3}b)}^2$

(vi) $\frac{9}{5}$s² + 6sv + 5v²
Taking out $\frac{9}{5}$ as a factor,
= $\frac{9}{5}(s^2+\frac{10}{3}sv+\frac{25}{9}{v}^2)$
= $\frac{9}{5}[(s{)}^2+2(s)\left(\frac{5}{3}v\right)+{\left(\frac{5}{3}v\right)}^2]$
Using a² + 2ab + b² = (a + b)²
= $\frac{9}{5}{(s+\frac{5}{3}v)}^2$
Therefore, $\frac{9}{5}$s² + 6sv + 5v² = = $\frac{9}{5}{(s+\frac{5}{3}v)}^2$

2. Find the values of the following using the identity (a – b)² = a² – 2ab + b².
(i) (79)²
(ii) (193)²
(iii) (299)²
Solution:
(i) (79)²
= (80 – 1)²
Using (a – b)² = a² – 2ab + b²
= (80)² – 2(80)(1) + (1)²
= 6400 – 160 + 1
= 6401 – 160
= 6241

(ii) (193)²
= (200 – 7)²
Using (a – b)² = a² – 2ab + b²
= (200)² – 2(200)(7) + (7)²
= 40000 – 2800 + 49
= 40049 – 2800
= 37249

(iii) (299)²
= (300 – 1)²
Using (a – b)² = a² – 2ab + b²
= (300)² – 2(300)(1) + (1)²
= 90000 – 600 + 1
= 90001 – 600
= 89401

Exercise Set 4.3

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) 117² (ii) 78² (iii) 198² (iv) 214² (v) 1104² (vi) 1120²
Solution:
(i) 117²
= (100 + 17)²
Using (a + b)² = a² + 2ab + b²
= (100)² + 2(100)(17) + (17)²
= 10000 + 3400 + 289
= 13689

(ii) 78²
= (70 + 8)²
Using (a + b)² = a² + 2ab + b²
= (70)² + 2(70)(8) + (8)²
= 4900 + 1120 + 64
= 6084

(iii) 198²
= (200 – 2)²
Using (a – b)² = a² – 2ab + b²
= (200)² – 2(200)(2) + (2)²
= 40000 – 800 + 4
= 40004 – 800
= 39204

(iv) 214²
= (200 + 14)²
Using (a + b)² = a² + 2ab + b²
= (200)² + 2 (200)(14) + (14)²
= 40000 + 5600 + 196
= 45796

(v) 1104²
= (1000 + 100 + 4)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (1000)² + (100)² + (4)² + 2(1000)(100) + 2(100)(4) + 2(4)(1000)
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816

(vi) 1120²
= (1000 + 100 + 20)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (1000)² + (100)² + (20)² + 2(1000)(100) + 2(100)(20) + 2(20)(1000)
= 1000000 + 10000 + 400 + 200000 + 4000 + 40000
= 1254400

2. Factor using suitable identities:
(i) 16y² – 24y + 9
(ii) $\frac{9}{4}$s² + 6st + 4t²
(iii) $\frac{m^2}{9}$ + $\frac{mk}{3}$ + $\frac{k^2}{4}$ + 3nk + 2mn + 9n²
(iv) $\frac{p^2}{16}$ – 2 + $\frac{16}{p^2}$
(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc
Solution:
(i) 16y² – 24y + 9
= (4y)² – 2(4y)(3) + (3)²
Using the identity a² – 2ab + b² = (a – b)²
= (4y – 3)²
Therefore, 16y² – 24y + 9 = (4y – 3)²

(ii) $\frac{9}{4}$s² + 6st + 4t²
= ${\left(\frac{3}{2}s\right)}^2$+ 2$\left(\frac{3}{2}s\right)$(2t) + $(2t{)}^2$
Using a² + 2ab + b² = (a + b)²
= ${(\frac{3}{2}s+2t)}^2$
Therefore, $\frac{9}{4}$s² + 6st + 4t² = ${(\frac{3}{2}s+2t)}^2$

(iii) $\frac{m^2}{9}$ + $\frac{mk}{3}$ + $\frac{k^2}{4}$ + 3nk + 2mn + 9n²
= $\frac{m^2}{9}$ + $\frac{k^2}{4}$ + 9n² + $\frac{mk}{3}$ + 3nk + 2mn
= ${\left(\frac{m}{3}\right)}^2$+ ${\left(\frac{k}{2}\right)}^2$+ $(3n{)}^2$ + 2$\left(\frac{m}{3}\right)$$\left(\frac{k}{2}\right)$ + 2$\left(\frac{k}{2}\right)$(3n) + 2(3n)$\left(\frac{m}{3}\right)$
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= ${(\frac{m}{3}+\frac{k}{2}+3n)}^2$
Therefore, $\frac{m^2}{9}$ + $\frac{mk}{3}$ + $\frac{k^2}{4}$ + 3nk + 2mn + 9n² = ${(\frac{m}{3}+\frac{k}{2}+3n)}^2$

(iv) $\frac{p^2}{16}$ – 2 + $\frac{16}{p^2}$
= ${\left(\frac{p}{4}\right)}^2$ – 2$\left(\frac{p}{4}\right)$$\left(\frac{4}{p}\right)$ + ${\left(\frac{4}{p}\right)}^2$
Using a² – 2ab + b² = (a – b)²
= ${\left(\frac{p}{4}\text{–}\frac{4}{p}\right)}^2$
Therefore, $\frac{p^2}{16}$ – 2 + $\frac{16}{p^2}$ = ${\left(\frac{p}{4}\text{–}\frac{4}{p}\right)}^2$

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc
= (3a)² + (–2b)² + c² + 2(3a)(–2b) + 2(3a)(c) + 2(–2b)(c)
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= [3a + (–2b) + c]²
= (3a – 2b + c)²
Therefore, 9a² + 4b² + c² − 12ab + 6ac − 4bc = (3a – 2b + c)²

3. Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)²
(ii) (3x – 2y + 4z)²
Solution:
(i) (p + 3q + 7r)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (p)² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
= p² + 9q² + 49r² + 6pq + 42qr + 14rp.

(ii) (3x – 2y + 4z)²
= [3x + (–2y) + 4z]²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (3x)² + (–2y)² + (4z)² + 2(3x)(–2y) + 2(–2y)(4z) + 2(4z)(3x)
= 9x² + 4y² + 16z² – 12xy – 16yz + 24xz.

4. Is this an identity?
(a + b − c)² + (a − b + c)² + (a − b − c)² = 2a + 2b + 2c .
Solution:
LHS = (a + b − c)² + (a − b + c)² + (a − b − c)²
Using (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
= [a² + b² + (–c)² + 2ab + 2a(–c) + 2b(–c)] + [a² + (–b)² + c² + 2a(–b) +2(–b)c + 2ac] + [a² + (–b)² + (–c)² + 2a(–b) + 2(–b)(–c) + 2a(–c)]
= (a² + b² + c² + 2ab − 2bc − 2ac) + (a² + b² + c² – 2ab − 2bc + 2ac) + (a² + b² + c² – 2ab + 2bc – 2ac)
= a² + a² + a² + b² + b² + b² + c² + c² + c² + 2ab – 2ab – 2ab – 2ac + 2ac – 2ac – 2bc – 2bc + 2bc
= 3a² + 3b² + 3c² – 2ab – 2bc – 2ac.
RHS = 2a + 2b + 2c.
Since LHS ≠ RHS
Hence, the given equation is not an identity.

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) s² – 11s + 24 = (________) (________)
s² – 11s + 24
= s² – 3s – 8s + 24
= s(s − 3) − 8(s − 3)
= (s − 3)(s − 8)
∴ s² – 11s + 24 = (s − 3)(s − 8)

(ii) (________) (x + 1) = (3x² – 4x – 7)
3x² – 4x – 7
= 3x² − 7x + 3x − 7
= x(3x − 7) + 1(3x − 7)
= (3x − 7)(x + 1)
∴ (3x − 7) (x + 1) = (3x² – 4x – 7)

(iii) 10x² – 11x – 6 = (2x – ___) (___ + 2)
10x² – 11x – 6
= 10x² − 15x + 4x − 6
= 5x(2x − 3) + 2(2x − 3)
= (2x − 3)(5x + 2)
∴ 10x² – 11x – 6 = (2x − 3)(5x + 2)

(iv) 6x² + 7x + 2 = (____________) (___________)
6x² + 7x + 2
= 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1)
∴ 6x² + 7x + 2 = (3x + 2)(2x + 1)

2. Select and use the identity that will help you to find the following products without multiplying directly: (i) (41)²
(ii) (27)²
(iii) (23 × 17)
(iv) (135)²
(v) (97)²
(vi) (18 × 29)
(vii) (34 × 43)
(viii) (205)²
Solution:
(i) (41)²
= (40 + 1)²
Using (a + b)² = a² + 2ab + b²
= (40)² + 2(40)(1) + (1)²
= 1600 + 80 + 1
= 1681.

(ii) (27)²
= (30 – 3)²
Using (a – b)² = a² – 2ab + b²
= (30)² – 2(30)(3) + (3)²
= 900 – 180 + 9
= 909 – 180
= 729.

(iii) (23 × 17)
= (20 + 3)(20 – 3)
Using (a + b)(a – b) = a² – b²
= (20)² – (3)²
= 400 – 9
= 391.

(iv) (135)²
= (100 + 30 + 5)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (100)² + (30)² + (5)² + 2(100)(30) + 2(30)(5) + 2(5)(100)
= 10000 + 900 + 25 + 6000 + 300 + 1000
= 18225.

(v) (97)²
= (100 – 3)²
Using (a – b)² = a² – 2ab + b²
= (100)² – 2(100)(3) + (3)²
= 10000 – 600 + 9
= 10009 – 600
= 9409.

(vi) (18 × 29)
= [{23 + (–5)} × (23 + 6)]
Using the identity (x + a)(x + b) = x² + (a + b)x + ab
= (23)² + {(–5) + 6}(23) + (–5)(6)
= 529 +(1)(23) + (–30)
= 529 + 23 – 30
= 552 – 30
= 522.

(vii) (34 × 43)
= {38 + (–4)} × (38 + 5)
Using (x + a)(x + b) = x² + (a + b)x + ab
= (38)² + {(–4) + 5}(38) + (–4)(5)
= 1444 +(1)(38) + (–20)
= 1444 + 38 – 20
= 1482 – 20
= 1462

(viii) (97)²
= (100 – 3)²
Using (a – b)² = a² – 2ab + b²
= (100)² – 2(100)(3) + (3)²
= 10000 – 600 + 9
= 9400 + 9
= 9409.

3. Factor the following:
(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc
(ii) 16s² + 25t² – 40st
(iii) r² – r – 42
(iv) 49g² + 14gh + h²
(v) 64u² + 121v² + 4w² – 176uv – 32uw + 44vw
Solution:
(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (3a)² + (–b)² + (2c)² + 2(3a)(–b) + 2(3a)(2c) + 2(–b)(2c)
= {3a + (–b) + 2c}²
= (3a – b + 2c)²

(ii) 16s² + 25t² – 40st
Using (a – b)² = a² – 2ab + b²
= (4s)² + (5t)² – 2(4s)(5t)
= (4s – 5t)²

(iii) r² – r – 42
= r² – 7r + 6r – 42
= r(r – 7) + 6(r – 7)
= (r – 7) (r + 6)

(iv) 49g² + 14gh + h²
Using (a + b)² = a² + 2ab + b²
= (7g)² + 2(7g)(h) + (h)²
= (7g + h)²

(v) 64u² + 121v² + 4w² – 176uv – 32uw + 44vw
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (–8u)² + (11v)² + (2w)² + 2(–8u)(11v) + 2(–8u)(2w) + 2(11v)(2w)
= (–8u + 11v + 2w)²

Exercise Set 4.5

1. Simplify the following rational expressions, assuming that the expressions in the denominators are not equal to zero:
(i) $\frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$
(ii) $\frac{n^3-3n^{2}m+3n{m}^2-m^3}{5m^2-10mn+5n^2}$
(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) $\frac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$
(vi) $\frac{p^4-16}{p^2-4p+4}$
Solution:
(i) $\frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$
Factorising the numerator:
$3p^2-3pq-18q^2$
= 3($p^2-pq-6q^2$)
= 3[p² – 3pq + 2pq – 6q²]
= 3[p(p – 3q) + 2q(p – 3q)]
= 3(p + 2q)(p – 3q)
Factorising the denominator:
$p^2+3pq-10q^2$
= p² + 5pq – 2pq – 10q²
= p(p + 5q) – 2q(p + 5q)
= (p + 5q)(p – 2q)
Now, since the denominator is not equal to zero.
Therefore,
$\frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$ = $\frac{3(p+2q)(p\text{–}3q)}{(p+5q)(p\text{–}2q)}$.

(ii) $\frac{n^3-3n^{2}m+3n{m}^2-m^3}{5m^2-10mn+5n^2}$
Factorising the numerator:
$n^3-3n^{2}m+3n{m}^2-m^3$
= (n)³ – 3(n)²(m) + 3(n)(m)² – (m)³
Using x³ – 3x²y + 3xy² – y³ = (x – y)³
= (n – m)³
Factorising the denominator:
$5m^2-10mn+5n^2$
= $5n^2-10mn+5m^2$
= 5($n^2-2mn+m^2$)
= 5[$(n{)}^2-2(m)(n)+(m{)}^2$]
Using a² – 2ab + b² = (a – b)²
= 5(n – m)²
Now, since the denominator is not equal to zero.
Therefore,
$\frac{n^3-3n^{2}m+3n{m}^2-m^3}{5m^2-10mn+5n^2}$ = $\frac{(n\text{–}m{)}^3}{5(n\text{–}m{)}^2}$ = $\frac{(n\text{–}m)}{5}$.

(iii) $\frac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}$
Factorising the numerator:
$w^3-v^3+x^3+3wvx$
Using the identity
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – xz – yz)
Thus,
$w^3-v^3+x^3+3wvx$ = w² + (–v)² + x² – 3w(–v)x = [w + (–v) + x] [(w)² + (–v)² + (x)² – (w)(–v) – (w)(x) – (–v)(x)]
⇒ $w^3-v^3+x^3+3wvx$ = (w – v + x)(w² + v² + x² + wv – wx + vx)
Factorising the denominator:
$w^2+v^2+x^2-2wv-2vx+2wx$
= (w)² + (–v)² + (x)² + 2(w)(–v) + 2(–v)(x) + 2(w)(x)
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= [w + (–v) + x]²
= (w – v + x)²
Now, since the denominator is not equal to zero.
Therefore,
$\frac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}$ = $\frac{(w\text{–}v+x)(w^2+v^2+x^2+wv\text{–}wx+vx)}{(w\text{–}v+x{)}^2}$ = $\frac{(w^2+v^2+x^2+wv\text{–}wx+vx)}{(w\text{–}v+x)}$.

(iv) $\frac{4y^2-20yz+25z^2}{(25z^2-4y^2)}$
Factorising the numerator:
$4y^2-20yz+25z^2$
= (2y)² – 2(2y)(5z) + (5z)²
Using (a – b)² = a² – 2ab + b²
= (2y – 5z)²
Factorising the denominator:
$25z^2-4y^2$
= $(5z{)}^2$ – $(2y{)}^2$
Using a² – b² = (a – b) (a + b)
= (5z – 2y) (5z + 2y)
Now, since the denominator is not equal to zero.
Therefore,
$\frac{4y^2-20yz+25z^2}{(25z^2-4y^2)}$ = $\frac{(2y\text{–}5z{)}^2}{(5z\text{–}2y)(5z+2y)}$ = $\frac{[\text{–}(5z\text{–}2y){]}^2}{(5z\text{–}2y)(5z+2y)}$ = $\frac{(5z\text{–}2y{)}^2}{(5z\text{–}2y)(5z+2y)}$ = $\frac{(5z\text{–}2y)}{(5z+2y)}$.

(v) $\frac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$
Factorising the numerator,
x² – x – 6 = x² + 3x – 2x – 6
= x(x + 3) – 2(x + 3)
= (x + 3) (x – 2)
x² – 7x + 12 = x² – 3x – 4x + 12
= x(x – 3) – 4(x – 3)
= (x – 3) (x – 4)
Factorising the denominator:
x² – 6x + 8 = x² – 2x – 4x + 8
= x(x – 2) – 4(x – 2)
= (x – 2) (x – 4)
x² – 9 = (x)² – (3)²
Using a² – b² = (a – b) (a + b)
= (x – 3) (x + 3)
Now, since the denominator is not equal to zero.
Therefore,
$\frac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$ = $\frac{(x+3)(x\text{–}2)(x\text{–}3)(x\text{–}4)}{(x\text{–}2)(x\text{–}4)(x\text{–}3)(x+3)}$ = 1.

(vi) $\frac{p^4-16}{p^2-4p+4}$
Factorising the numerator,
p⁴ – 16 = (p²)² – (4)²
Using a² – b² = (a – b) (a + b)
= (p² – 4) (p² + 4)
= [(p)² – (2)²] (p² + 4)
= (p – 2) (p + 2) (p² + 4)
Factorising the denominator,
p² – 4p + 4 = p² – 2p – 2p + 4
= p(p – 2) – 2(p – 2)
= (p – 2) (p – 2)
Therefore,
$\frac{p^4-16}{p^2-4p+4}$ = $\frac{(p\text{–}2)(p+2)(p^2+4)}{(p\text{–}2)(p\text{–}2)}$ = $\frac{(p+2)(p^2+4)}{(p\text{–}2)}$.

End of Chapter Exercise (Page 88 – 90)

1. Use suitable identities to find the following products:
(i) (–3x + 4)²
(ii) (2s + 7) (2s – 7)
(iii) $(p^2+\frac{1}{2})(p^2-\frac{1}{2})$
(iv) (2n + 7) (2n – 7)
(v) (s – 2t) (s² + 2st + 4t²)
(vi) ${(\frac{1}{2r}-4r)}^2$
(vii) (–3m + 4k – l)²
(viii) ${(x-\frac{1}{3}y)}^3$
(ix) ${(\frac{7}{2}k-\frac{2}{3}m)}^3$
Solution:
(i) (–3x + 4)²
Using (a + b)² = a² + 2ab + b²
= (–3x)² + 2(–3x)(4) + (4)²
= 9x² – 24x + 16

(ii) (2s + 7) (2s – 7)
Using (a + b) (a – b) = a² – b²
= (2s)² – (7)²
= 4s² – 49

(iii) $(p^2+\frac{1}{2})(p^2-\frac{1}{2})$
Using (a + b) (a – b) = a² – b²
= (p²)² – ${\left(\frac{1}{2}\right)}^2$
= p⁴ – $\frac{1}{4}$

(iv) (2n + 7) (2n – 7)
Using (a + b) (a – b) = a² – b²
= (2n)² – (7)²
= 4n² – 49

(v) (s – 2t) (s² + 2st + 4t²)
Using (a − b)(a² + ab + b²) = a³ − b³
= (s – 2t) [(s)² + (s)(2t) + (2t)²]
= (s)³ – (2t)³
= s³ – 8t³

(vi) ${(\frac{1}{2r}-4r)}^2$
Using (a – b)² = a² – 2ab + b²
= ${\left(\frac{1}{2r}\right)}^2$– 2$\left(\frac{1}{2r}\right)$(4r) + (4r)²
= $\frac{1}{4r^2}$ – 4 + 16r²
= 16r² + $\frac{1}{4r^2}$ – 4

(vii) (–3m + 4k – l)²
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
= (–3m)² + (4k)² + (–l)² + 2(–3m)(4k) + 2(4k)(–l) + 2(–l)(–3m)
= 9m² + 16k² + l² – 24mk – 8kl + 6ml

(viii) ${(x-\frac{1}{3}y)}^3$
Using (a – b)³ = a³ – 3a²b + 3ab² – b³
= (x)³ – 3(x)²$\left(\frac{1}{3}y\right)$ + 3(x)${\left(\frac{1}{3}y\right)}^2$ – ${\left(\frac{1}{3}y\right)}^3$
= x³ – 3.x².$\frac{1}{3}y$ + 3.x.$\frac{1}{9}{y}^2$ – $\frac{1}{27}{y}^3$
= x³ – x²y + $\frac{1}{3}$xy² – $\frac{1}{27}{y}^3$

(ix) ${(\frac{7}{2}k-\frac{2}{3}m)}^3$
Using (a – b)³ = a³ – 3a²b + 3ab² – b³
= ${\left(\frac{7}{2}k\right)}^3$– 3${\left(\frac{7}{2}k\right)}^2$$\left(\frac{2}{3}m\right)$ + 3$\left(\frac{7}{2}k\right)$${\left(\frac{2}{3}m\right)}^2$ – ${\left(\frac{2}{3}m\right)}^3$

= $\frac{343}{8}{k}^2$ – 3. $\frac{49}{4}{k}^2$.$\frac{2}{3}m$ + 3.$\frac{7}{2}k$.$\frac{4}{9}{m}^2$ – $\frac{8}{27}{m}^3$

= $\frac{343}{8}{k}^2$ – $\frac{49}{2}{k}^2$m + $\frac{14}{3}k$$m^2$ – $\frac{8}{27}{m}^3$

2. Find the values using suitable identities:
(i) 17 × 21
(ii) 104 × 96
(iii) 24 × 16
(iv) 147³
(v) 199³
(vi) 127³
(vii) (–107)³
(viii) (–299)³
Solution:
(i) 17 × 21
Using (a – b) (a + b) = a² – b²
= (19 – 2) (19 + 2)
= (19)² – (2)²
= 361 – 4
= 357

(ii) 104 × 96
Using (a + b) (a – b) = a² – b²
= (100 + 4) (100 – 4)
= (100)² – (4)²
= 10000 – 16
= 9984

(iii) 24 × 16
Using (a + b) (a – b) = a² – b²
= (20 + 4) (20 – 4)
= (20)² – (4)²
= 400 – 16
= 384

(iv) 147³
Using (a – b)³ = a³ – 3a²b + 3ab² – b³
= (150 – 3)³
= (150)³ – 3(150)²(3) + 3(150)(3)² – (3)³
= 3375000 − 202500 + 4050 − 27
= 3176523

(v) 199³
Using (a – b)³ = a³ – 3a²b + 3ab² – b³
= (200 – 1)³
= (200)³ – 3(200)²(1) + 3(200)(1)² – (1)³
= 8000000 − 120000 + 600 − 1
= 7880599

(vi) 127³
Using (a + b)³ = a³ + 3a²b + 3ab² + b³
= (100 + 27)³
= (100)³ – 3(100)²(27) + 3(100)(27)² – (27)³
= 1000000 + 810000 + 218700 + 19683
= 2048383

(vii) (–107)³
Using (a + b)³ = a³ + 3a²b + 3ab² + b³
= (–100 – 7)³
= (–100)³ + 3(–100)²(–7) + 3(–100)(–7)² + (–7)³
= −1000000 − 210000 − 14700 − 343
= −1225043

(viii) (–299)³
Using (a + b)³ = a³ + 3a²b + 3ab² + b³
= (–300 + 1)³
= (–300)³ + 3(–300)²(1) + 3(–300)(1)² + (1)³
= −27000000 + 270000 − 900 + 1
= −26730900 + 1
= −26730899

3. Factor the following algebraic expressions:
$(i)4y^2+1+\frac{1}{16y^2}$
$(ii)9m^2-\frac{1}{25n^2}$
$(iii)27b^3-\frac{1}{64b^3}$
$(iv)x^2+\frac{5x}{6}+\frac{1}{6}$
$(v)27u^3-\frac{1}{125}-\frac{27u^2}{5}+\frac{9u}{25}$
$(vi)64y^3+\frac{1}{125}{z}^3$
$(vii)p^3+27q^3+r^3-9pqr$
$(viii)9m^2-12m+4$
$(ix)9x^3-\frac{8}{3}{y}^3+\frac{z^3}{3}+6xyz$
$(x)4x^2+9y^2+36z^2+12xz+36yz+24xy$
$(xi)27u^3-\frac{1}{216}-\frac{9u^2}{2}+\frac{u}{4}$
Solution:
(i) $4y^2+1+\frac{1}{16y^2}$
Using a² + 2ab + b² = (a + b)²
= (2y)² + 2(2y)$\left(\frac{1}{4y}\right)$ + ${\left(\frac{1}{4y}\right)}^2$
= ${(2y+\frac{1}{4y})}^2$

(ii) $9m^2-\frac{1}{25n^2}$
Using a² – b² = (a – b) (a + b)
= (3m)² – ${\left(\frac{1}{5n}\right)}^2$
= $\left(3m\text{–}\frac{1}{5n}\right)$ $(3m+\frac{1}{5n})$

(iii) $27b^3-\frac{1}{64b^3}$
Using a³ − b³ = (a − b)(a² + ab + b²)
= (3b)³ – ${\left(\frac{1}{4b}\right)}^3$
= $\left(3b\text{–}\frac{1}{4b}\right)$ $[(3b{)}^2+(3b)\left(\frac{1}{4b}\right)+{\left(\frac{1}{4b}\right)}^2]$
= $\left(3b\text{–}\frac{1}{4b}\right)$$(9b^2+\frac{3}{4}+\frac{1}{16b^2})$

(iv) $x^2+\frac{5x}{6}+\frac{1}{6}$
= x² + $\frac{x}{3}$ + $\frac{x}{2}$ + $\frac{1}{6}$
= x$(x+\frac{1}{3})$+ $\frac{1}{2}$$(x+\frac{1}{3})$
= $(x+\frac{1}{3})$ $(x+\frac{1}{2})$

(v) $27u^3-\frac{1}{125}-\frac{27u^2}{5}+\frac{9u}{25}$
Using a³ – 3a²b + 3ab² – b³ = (a – b)³
= (3u)³ – ${\left(\frac{1}{5}\right)}^3$ – 3(3u)²$\left(\frac{1}{5}\right)$ + 3(3u)${\left(\frac{1}{5}\right)}^2$
= ${\left(3u\text{–}\frac{1}{5}\right)}^3$

(vi) $64y^3+\frac{1}{125}{z}^3$
Using a³ + b³ = (a + b)(a² – ab + b²)
= (4y)² + ${\left(\frac{z}{5}\right)}^3$
= $(4y+\frac{z}{5})$ $[(4y{)}^2\text{–}(4y)\left(\frac{z}{5}\right)+{\left(\frac{z}{5}\right)}^2]$
= $(4y+\frac{z}{5})$$(16y^2\text{–}\frac{4yz}{5}+\frac{z^2}{25})$

(vii) $p^3+27q^3+r^3-9pqr$
Using x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – xz – yz)
$p^3+27q^3+r^3-9pqr$ = (p)³ + (3q)³ + (r)³ – 3(p)(3q)(r)
Thus,
x = p, y = 3q, z = r
So,
$p^3+27q^3+r^3-9pqr$ = (p + 3q + r)[(p)² + (3q)² + (r)² – (p)(3q) – (3q)(r) – (p)(r)]
= (p + 3q + r)(p² + 9q² + r² – 3pq – pr – 3qr)

(viii) $9m^2-12m+4$
Using a² – 2ab + b² = (a – b)²
= (3m)² – 2(3m)(2) + (2)²
= (3m – 2)²

(ix) $9x^3-\frac{8}{3}{y}^3+\frac{z^3}{3}+6xyz$
Taking out $\frac{1}{3}$ as a factor,
= $\frac{1}{3}$$(27x^3\text{–}8y^3+z^3+18xyz)$
Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
$27x^3\text{–}8y^3+z^3+18xyz$ = (3x)³ + (–2y)³ + (z)³ – 3(3x)(–2y)(z)
Thus,
a = 3x, b = (–2y) and c = z
$\frac{1}{3}$$(27x^3\text{–}8y^3+z^3+18xyz)$ = $\frac{1}{3}$ [3x + (–2y) + z] [(3x)² + (–2y)² + (z)² – (3x)(–2y) – (–2y)(z) – (3x)(z)]
= $\frac{1}{3}$ (3x –2y + z) (9x² + 4y² + z² + 6xy + 2yz – 3xz)

(x) $4x^2+9y^2+36z^2+12xz+36yz+24xy$
Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)²

(xi) $27u^3-\frac{1}{216}-\frac{9u^2}{2}+\frac{u}{4}$
Using a³ – 3a²b + 3ab² – b³ = (a – b)³
= (3u)³ – ${\left(\frac{1}{6}\right)}^3$ – 3(3u)²$\left(\frac{1}{6}\right)$ + 3(3u)${\left(\frac{1}{6}\right)}^2$
= ${\left(3u\text{–}\frac{1}{6}\right)}^3$

4. Simplify the following:
$(i)\frac{4x^2+4x+1}{4x^2-1}$
$(ii)\frac{9(3a^3-24b^3)}{9a^2-36b^2}$
$(iii)\frac{s^3+125t^3}{s^2-2st-35t^2}$
Note: Assume that the denominators are not equal to 0.
Solution:
(i) $\frac{4x^2+4x+1}{4x^2-1}$
Factorising the numerator:
$4x^2+4x+1$
Using a² + 2ab + b² = (a + b)²
= (2x)² + 2(2x)(1) + (1)²
= (2x + 1)²
Factorising the denominator:
$4x^2-1$
Using a² – b² = (a – b) (a + b)
= (2x)² – (1)²
= (2x – 1) (2x + 1)
Since the denominator is not equal to zero.
Therefore,
$\frac{4x^2+4x+1}{4x^2-1}$ = $\frac{(2x+1{)}^2}{(2x\text{–}1)(2x+1)}$ = $\frac{(2x+1)}{(2x\text{–}1)}$

(ii) $\frac{9(3a^3-24b^3)}{9a^2-36b^2}$
= $\frac{9\times 3(a^3-8b^3)}{9(a^2-4b^2)}$
= $\frac{3(a^3-8b^3)}{(a^2-4b^2)}$
Factorising the numerator:
$3(a^3-8b{)}^3$
Using a³ − b³ = (a − b)(a² + ab + b²)
= 3[(a)³ – (2b)]³
= 3[(a − 2b) {a² + (a)(2b) + (2b)²}]
= 3(a – 2b) (a² + 2ab + 4b²)
Factorising the denominator:
$a^2-4b^2$
Using a² – b² = (a – b) (a + b)
= (a)² – (2b)²
= (a – 2b) (a + 2b)
Since the denominator is not equal to zero.
Therefore,
$\frac{3(a^3-8b^3)}{(a^2-4b^2)}$ = $\frac{3(a\text{–}2b)(a^2+2ab+4b^2)}{(a\text{–}2b)(a+2b)}$ = $\frac{3(a^2+2ab+4b^2)}{(a+2b)}$

(iii) $\frac{s^3+125t^3}{s^2-2st-35t^2}$
Factorising the numerator:
$s^3+125t^3$
Using a³ + b³ = (a + b)(a² – ab + b²)
= (s)³ + (5t)³
= (s + 5t)$[(s{)}^2\text{–}(s)\left(5t\right)+{\left(5t\right)}^2]$
= (s + 5t) (s² – 5st + 25t²)
Factorising the denominator:
$s^2-2st-35t^2$
= s² – 7st + 5st – 35t²
= s(s – 7t) + 5t(s – 7t)
= (s – 7t) (s + 5t)
Since the denominator is equal to zero.
Therefore,
$\frac{s^3+125t^3}{s^2-2st-35t^2}$ = $\frac{(s+5t)(s^2\text{–}5st+25t^2)}{(s\text{–}7t)(s+5t)}$ = $\frac{(s^2\text{–}5st+25t^2)}{(s\text{–}7t)}$

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a² – 30ab + 9b²
(ii) 36s² – 49t²
Solution:
(i) Area = 25a² – 30ab + 9b²
Using a² − 2ab + b² = (a − b)²
25a² − 30ab + 9b² = (5a)² − 2(5a)(3b) + (3b)²
= (5a − 3b)²
Hence, possible expressions for the length and breadth are (5a − 3b) and (5a − 3b).​

(ii) Area = 36s² – 49t²
Using a² – b² = (a – b) (a + b)
36s² – 49t² = (6s)² – (7t)²
= (6s – 7t) (6s + 7t)
Hence, possible expressions for the length and breadth are (6s − 7t) and (6s + 7t).​

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a² – 24b²
(ii) 3ps² – 15ps + 12p
Solution:
(i) Volume = 6a² – 24b²
= 6 (a² – 4b²)
Using a² – b² = (a – b) (a + b)
= 6 [(a)² – (2b)²]
= 6 (a – 2b) (a + 2b)
Therefore,
6a² – 24b² = 6 (a – 2b) (a + 2b)
Hence, possible expressions for the length, breadth, and height are 6, (a − 2b), (a + 2b).

(ii) Volume = 3ps² – 15ps + 12p
= 3p (s² – 5s + 4)
= 3p [s² – 4s – s + 4]
= 3p [s(s – 4) –1(s – 4)]
= 3p (s – 4) (s – 1)
Therefore,
3ps² – 15ps + 12p = 3p (s – 4) (s – 1)
Hence, possible expressions for the length, breadth, and height are 3p, (s – 4), (s – 1).

7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:

Side of the playground = 40 m
Width of the path = s m
Side of outer sqaure = (40 + 2s) m
Area of the outer square = (40 + 2s)² m²
Area of the square = (40)² m²
Area of path = Outer area – Inner area
= (40 + 2s)² – (40)²
Using (a + b)² = a² + 2ab + b²
= (40)² + 2(40)(2s) + (2s)² – (40)²
= 1600 + 160s + 4s² – 1600
= 160s + 4s²
Hence, the required expression for the area of the path is (4s² + 160s)​ m².

8. If a number plus its reciprocal equals $\frac{10}{3}$ , find the number.
Solution:
Let the number be x.
Then,
x + $\frac{1}{x}$ = $\frac{10}{3}$
$\frac{x^2+1}{x}$ = $\frac{10}{3}$
By cross multiplication,
3(x² + 1) = 10x
3x² + 3 = 10x
3x² – 10x + 3 = 0
3x² – 9x – x + 3 = 0
3x(x – 3) –1 (x – 3) = 0
(x – 3) (3x – 1) = 0
x = 3 and x = $\frac{1}{3}$
Hence, the numbers are 3 and $\frac{1}{3}$.

9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Solution:
Area of the pool = 2x² + 7x + 3
Width of the pool = 2x + 1
Length = $\frac{Area}{Breadth}$ = $\frac{2x^2+7x+3}{2x+1}$
Factorising the numerator:
2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3) (2x + 1)
Therefore,
Length = $\frac{(x+3)(2x+1)}{(2x+1)}$ = (x + 3)
Hence, the length of the rectangular pool is (x + 3) hastas.

10. If both x – 2 and x – $\frac{1}{2}$ are factors of px² + 5x + r, show that p = r.

11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25.
Solution:
Using the identity,
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² + 2ab + 2bc + 2ac – 3ab – 3bc – 3ca)
a³ + b³ + c³ – 3abc = (a + b + c) [(a + b + c)² – 3(ab + bc + ca)]
On substituting a + b + c = 5 and ab + bc + ca = 10,
a³ + b³ + c³ – 3abc = (5) [(5)² – 3(10)]
a³ + b³ + c³ – 3abc = (5)(25 – 30)
a³ + b³ + c³ – 3abc = (5)(– 5)
a³ + b³ + c³ – 3abc = (–25)
Hence, proved.

12. By factoring the expression, check that n³ – n is always divisible by 6 for all natural numbers n. Give reasons.
Solution:
n³ – n = n(n² – 1)
= n × [(n)² – (1)²]
Using a² – b² = (a – b) (a + b)
= n × (n – 1) × (n + 1)
Thus,
n³ – n = n(n – 1)(n + 1)
This is the product of three consecutive natural numbers.
Among any three consecutive natural numbers:
• one number is always divisible by 3,
• at least one number is always even, i.e., divisible by 2.
Therefore, the product n(n − 1)(n + 1) is divisible by both 2 and 3.
Hence, it is divisible by 2 × 3 = 6.
Therefore,
n³ − n is always divisible by 6 for all natural numbers n.

13. Find the value of
(i)  x³ + y³ – 12xy + 64, when x + y = – 4
(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6
Solution:
(i)  Given that x + y = – 4
x + y + 4 = 0 …….. (1)
x³ + y³ – 12xy + 64 = x³ + y³ + (4)³ – 3(x)(y)(4)
Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ac)
x³ + y³ + 4³ – 3(x)(y)(4) = (x + y + 4)[x² + y² + (4)² – xy – (y)(4) –(x)(4)]
x³ + y³ – 12xy + 64 = (x + y + 4)(x² + y² + 16 – xy – 4y –4x)
x³ + y³ – 12xy + 64 = 0 × (x² + y² + 16 – xy – 4y –4x) …………. [Using (1)]
x³ + y³ – 12xy + 64 = 0
Hence, the value is 0.

(ii) Given that x = 2y + 6
x – 2y – 6 = 0 ……… (1)
x³ – 8y³ – 36xy – 216 = x³ + (–2y)³ + (–6)³ – 3(x)(–2y)(–6)
Using a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ac)
x³ + (–2y)³ + (–6)³ – 3(x)(–2y)(–6) = [x + (–2y) + (–6)] [x² + (–2y)² + (–6)² – (x)(–2y) – (–2y)(–6) –(x)(–6)]
x³ – 8y³ – 36xy – 216 = (x – 2y – 6) (x² + 4y² + 36 + 2xy – 12y + 6x)
x³ – 8y³ – 36xy – 216 = 0 × (x² + 4y² + 36 + 2xy – 12y + 6x) …………. [Using (1)]
x³ – 8y³ – 36xy – 216 = 0
Hence, the value is 0.

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