Algebraic Identities ICSE Class-8th Concise Selina

Algebraic Identities ICSE Class-8th Concise Selina Mathematics Solutions Chapter-12 . We provide step by step Solutions of Exercise / lesson-12 Algebraic Identities for ICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-12 A,  Exe-12 B, Exe-12 C  and Exe-12 D to develop skill and confidence . Visit official Website CISCE for detail information about ICSE Board Class-8.

Algebraic Identities ICSE Class-8th Concise Selina Mathematics  Solutions Chapter-12


–: Select Topics :–

 

Exe-12 A,

Exe-12 B,

Exe-12 C,

Exe-12 D,

 


ICSE Class-8  Algebraic Identities Exe-12 A,

Question 1 :-

Use direct method to evaluate the following products :
(i) (x + 8)(x + 3)
(ii) (y + 5)(y – 3)
(iii) (a – 8)(a + 2)
(iv) (b – 3)(b – 5)
(v) (3x – 2y)(2x + y)
(vi) (5a + 16)(3a – 7)
(vii) (8 – b) (3 + b)
Answer :-

(i)

(x + 8) (x + 3)

= (x × x) + (x × 3) + (8 × x) + (8 × 3)

= x2 + 3x + 8x + 24

= x2 + 11x + 24

(ii)

(y + 5) (y – 3) = (y × y) + (y × −3) + (5 × y) + (5 × −3)

= y2 + (−3y) + (5y) − 15

= y− 3y + 5y − 15

= y2 + 2y − 15

(iii)

(a – 8) (a + 2) = (a × a) + (a × 2) + (−8)×a + (−8)(2)

= a+ 2a − 8a − 16

= a2 − 6a − 16

(iv)

(b – 3) (b – 5) = (b × b) + (b × −5) + (−3) × b + (−3) (−5)

= b2 − 5b − 3b + 15

= b− 8b + 15

(v)

(3x – 2y) (2x + y) = (3x × 2x) + (3x × y) + (−2y × 2x) + (−2y × y)

= 6x2 + 3xy − 4xy − 2y

= 6x2 − xy − 2y2

(vi)

(5a + 16) (3a – 7) = (5a × 3a) + (5a × −7) + (16 ×  3a) + 16 −7

= 15a2 + (−35a) + 48a + (−112)

= 15a− 35a + 48a − 112

= 15a2 + 13a − 112

(vii)

(8 – b) (3 + b) = (8 × 3) + (8 × b) + (−b × 3) + (−b × b)

= 24 + 8b − 3b − b2

= 24 + 5b − b2

Question 2 :-

Use direct method to evaluate :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-A-2
Answer :-

(i)

(a+b) (a−b) = a2 − b2

(x+1) (x−1) = (x)2 − (1)2

= x− 1

(ii)

(a+b) (a−b) = a2 − b2

(2+a) (2−a) = (2)2 − (a)2

= 4 − a2

(iii)

(a+b) (a−b) = a2 − b2

(3b−1) (3b+1)

= (3b)− (1)2

= 9b2 − 1

(iv)

(a+b) (a−b) = a2 − b2

(4+5x) (4−5x)

= (4)2 − (5x)2

= 16 − 25x2

(v)

(a+b) (a−b) = a2 − b2

(2a+3) (2a−3) = (2a)2 − (3)2

= 4a− 9

(vi)

(a+b) (a−b) = a2 − b2

(xy+4) (xy−4) = (xy)2 − (4)2

= x2y2 − 16

(vii)

(a+b) (a−b) = a2 − b2

(ab+x2) (ab−x2) = (ab)2 − (x2)2

= a2b2 − x4

(viii)

(a+b) (a−b) = a2 − b2

(3x2+5y2) (3x2−5y2) = (3x2)2 − (5y2)2

= 9x4 − 25y4

(ix)

(a+b) (a−b) = a2 − b2

ICSE Class-8 Algebraic Identities Exe-12 A, img 1

(x) 

(a+b) (a−b) = a2 − b2

ICSE Class-8 Algebraic Identities Exe-12 A, img 2

(xi)

(a+b) (a−b) = a2 − b2

(0.5a−2a) (0.5+2a)

= (0.5)2 − (2a)2

= 0.25 − 4a2

(xii) 

(a+b) (a−b) = a2 − b2

ICSE Class-8 Algebraic Identities Exe-12 A, img 3

Question 3 :-

Evaluate :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-A-3
Answer :-

(i)

(a+1) (a-1) (a2+1)

= [(a)2−(1)2] (a2+1)

= (a2−1) (a2+1)

= (a2)2 − (1)2

= a4 − 1

(ii)

(a+b) (a−b) (a2+b2)

= (a2−b2) (a2+b2)

= (a2)− (b2)2

= a4 − b4

(iii)

(2a−b) (2a+b) (4a2+b2)

= [(2a)2−(b)2] (4a2+b2)

= (4a2−b2) (4a2+b2)

= (4a2)2 − (b2)2

= 16a4 − b4

(iv)

(3−2x) (3+2x) (9+4x2)

= [{3}2−(2x)2] (9+4x2)

= (9−4x2) (9+4x2)

= (9)2 − (4x2)2

= 81 − 16x4

(v)

(3−2x) (3+2x) (9+4x2)

= [{3}2−(2x)2] (9+4x2)

= (9−4x2) (9+4x2)

= (9)2 − (4x2)2

= 81 − 16x4

Question 4 :-

Use the product (a + b) (a – b) = a2 – b2 to evaluate:
(i) 21 × 19
(ii) 33 × 27
(iii) 103 × 97
(iv) 9.8 × 10.2
(v) 7.7 × 8.3
(vi) 4.6 × 5.4
Answer :-

(i)

21 x 19 = (20+1) (20−1)

= (20)2 − (1)2

= 400 − 1

= 399

(ii)

33 x 27 = (30+3) (30−3)

= (30)2 − (3)2

= 900 − 9

= 891

(iii)

103 x 97 = (100+3) (100−3)

= (100)2 − (3)2

= 10000 − 9

= 9991

(iv)

9.8 x 10.2 = (10−0.2) (10+0.2)

= (10)2 − (0.2)

= 100 − 0.04

= 99.96

(v)

7.7 x 8.3 = (8−0.3) (8+0.3)

= (8)− (0.3)

= 64 − 0.09

= 63.91

(vi)

4.6 x 5.4 = (5−0.4) (5+0.4)

= (5)− (0.4)2

= 25 − 0.16

= 24.84

Question 5 :-

Evaluate :
(i) (6 – xy) (6 + xy)
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-A-5
Answer :-

(i)

(6 – xy) (6 + xy) = 6(6+xy) − xy(6+xy)

= 36 + 6xy − 6xy + (xy)2

= 36 − x2 − y2

(ii) ICSE Class-8 Algebraic Identities Exe-12 A, img 4

(iii) ICSE Class-8 Algebraic Identities Exe-12 A, img 5

(iv) ICSE Class-8 Algebraic Identities Exe-12 A, img 6

(v)

(2a +3) (2a − 3) (4a2 + 9)

= [(2a)2 − (3)2] (4a2 + 9) ……….[(a+b) (a−b) = a2 − b2]

= (4a2 − 9) (4a2 + 9)

= (4a2)2 − (9)2 ………[(a+b) (a−b) = a2 − b2]

= 16a4 − 81

(vi)

(a + bc) (a − bc) (a2 + b2c2)

= [(a)2 − (bc)2] (a2 + b2c2)  ………..[(a+b) (a−b) = a2 − b2]

= (a− b2c2) (a+ b2c2) ……….[ ∵ (a+b) (c−b) = a2 − b2]

= a4 − b4c4

(vii)

(5x + 8y) (3x + 5y)

= 5x (3x + 5y) + 8y (3x+ 5y)

= 15x2 + 25xy + 24xy + 40y2

= 15x2 + 49xy + 40y2

(viii)

(7x + 15y) (5x − 4y)

= 7x (5x − 4y) + 15y (5x − 4y)

= 35x2 − 28xy + 75xy − 60y2

= 35x2 + 47xy − 60y2

(ix)

(2a − 3b) (3a + 4b)

= 2a (3a + 4b) − 3b (3a + 4b)

= 6a2 + 8ab − 9ab − 12b2

= 6a2 − ab − 12b2

(x)

(9a − 7b) (3a − b)

= 9a (3a − b) − 7b (3a − b)

= 27a2 − 9ab − 21ab + 7b2

= 27a2 − 30ab + 7b2


Exercise – 12 B Algebraic Identities ICSE Class-8th Mathematics 

Question 1 :-

Expand :
(i) (2a + b)2
(ii) (a – 2b)2
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-B-1
Answer :-

(i)

(2a + b)= (2a)2 + (b)2 + 2 × 2a × b ……[(a+b)2 = a2 + b+ 2ab]

= 4a2 + b2 + 4ab

(ii)

(a – 2b)2 = (a)2 + (2b)− 2 × a × 2b ……..[(a−b)2 = a2 + b− 2ab]

= a+ 4b− 4ab

(iii) ICSE Class-8 Algebraic Identities Exe-12 B, img 7

(iv) ICSE Class-8 Algebraic Identities Exe-12 B, img 8

(v)

(a+b−c)2 = (a)2 + (b)2 +(−c)2 + 2 × a × b + 2 × b × (−c) + 2 × (−c) × (a)

= a2 + b2 + c2 + 2ab − 2bc − 2ca

(vi)

(a+b+c)= a2 + b2 + c2 + 2ab − 2bc − 2ca

 (a−b+c)2 = (a)2 + (−b)2 + (c)2 + 2 × a × −b + 2(−b)(c) + 2× c × a

= a+ b+ c2 − 2ab − 2bc + 2ca

(vii) ICSE Class-8 Algebraic Identities Exe-12 B, img 9

(viii) ICSE Class-8 Algebraic Identities Exe-12 B, img 10

Question 2 :-

Find the square of :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-B-2
Answer :-

(i)

(x + 3y)2 = (x)2 + (3y)+ 2× x × 3y

= x2 + 9y+ 6xy

(ii)

(2x − 5y)2 = (2x)2 + (5y)2 − 2 × 2x × 5y

= 4x2 + 25y2 − 20xy

(iii) ICSE Class-8 Algebraic Identities Exe-12 B, img 11

(iv) ICSE Class-8 Algebraic Identities Exe-12 B, img 12

(v)

(x − 2y +1)2 = (x)2 + (−2y)+ (1)2 + 2 × x × −2y + 2 × (−2y) ×1 + 2 × 1 × x

= x2 + 4y2 + 1 − 4xy − 4y + 2x

(vi)

(3a − 2b − 5c)2 = (3a)2 + (−2b)+ (−5c)2 + 2 × 3a × −2b + 2 × (−2b)(−5c) + 2 × −5c × 3a

= 9a2 + 4b2 + 25c− 12ab + 20bc − 30ca

(vii) ICSE Class-8 Algebraic Identities Exe-12 B, img 13

(viii) ICSE Class-8 Algebraic Identities Exe-12 B, img 14

(ix)

(2x − 3y + z)2 = (2x)2 + (−3y)+ (z)2 + 2 × 2x × −3y + 2(−3y) × z + 2 × z × 2x

= 4x2 + 9y+ z− 12xy − 6yz + 4zx

(x) ICSE Class-8 Algebraic Identities Exe-12 B, img 15

Question 3 :-

Evaluate:
Using expansion of (a + b)2 or (a – b)2
(i) (208)2
(ii) (92)2
(iii)(415)2
(iv) (188)2
(v) (9.4)2
(vi) (20.7)2
Answer :-

(i)

(208)2 = (200 + 8)2

= (200)+ (8)2 + 2(200) (8)

= 40000 + 64 + 3200

= 43264

(ii)

(92)2 = (100 − 8)

= (100)+ (8)2 − 2(100) (8)

= 10000 + 64 − 1600

= 10064 − 1600

= 8464

(iii)

(415)2 = (400 + 15)2

= (400)+ (15)+ 2(400)(15)

= 160000 + 225 + 12000

= 172225

(iv)

(188)2 = (200 − 12)

= (200)+ (12)2 − 2(200) (12)

= 40000 + 144 − 4800

= 40144 − 4800

= 35344

(v)

(9.4)= (10 − 0.6)2

= (10)2 + (0.6)2 − 2 (10) (0.6)

= 100 + 0.36 − 12

= 88 + 0.36

= 88.36

(vi)

(20.7)2 = (20 + 0.7)2

= (20)2 + (0.7)2 + 2 (20) (0.7)

= 400 + 0.49 + 28

= 428 + 0.49

= 428.49

Question 4 :-

Expand :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-B-4
Answer :-

(i)

(2a + b)3 = (2a)3 + (b)3 + 3×2a×b(2a + b) ……[(a + b)3 = a+ b3 + 3ab(a + b)]

= 8a3 + b3 + 6ab (2a + b)

= 8a+ b+ 12a2b + 6ab2

(ii)

(a − 2b)3 = (a)− (2b)− 3×a×2b (a − 2b) ………[(a − b)= a− b− 3ab(a − b)]

= a− 8b− 6ab (a − 2b)

= a3 − 8b3 − 6a2b + 12ab2

(iii)

(3x − 2y)2 = (3x)− (2y)− 3 × 3x × 2y (3x − 2y)

= 27x− 8y3 − 18xy (3x − 2y)

= 27x− 8y− 54x2y + 36xy2

(iv)

(3x − 2y)2 = (3x)− (2y)− 3 × 3x × 2y (3x − 2y)

= 27x− 8y3 − 18xy (3x − 2y)

= 27x− 8y− 54x2y + 36xy2

(v)ICSE Class-8 Algebraic Identities Exe-12 B, img 16

(vi)ICSE Class-8 Algebraic Identities Exe-12 B, img 17

Question 5 :-

Find the cube of :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-B-5
Answer :-

(i)

(a + 2)3 = (a)+ (2)+ 3×a×2(a + 2)

= a3 + 8 + 6a(a+ 2)

= a+ 8 + 6a+ 12a

= a+ 6a2 + 12a+ 8

(ii)

(2a − 1)3 = (2a)3 − (1)3 − 3×2a×1(2a − 1)

= 8a− 1 − 6a (2a − 1)

= 8a3 − 1 − 12a2 + 6a

= 8a3 − 12a2 + 6a − 1

(iii)

(2a + 3b)3 = (2a)3 + (3b)+ 3×2a×3b(2a + 3b)

= 8a3 + 27b3 + 18ab (2a + 3b)

= 8a+ 27b3 + 36a2b + 54ab2

= 8a+ 36a2b + 54ab2 + 27b3

(iv)

(3b − 2a)3 = (3b)3 − (2a)3 − 3×3b×2a(3b − 2a)

= 27b3 − 8a3 − 18ab (3b − 2a)

= 27b− 8a3 − 54ab2 + 36a2b

= 27b− 54b2a + 36ba2 − 8a3

(v)ICSE Class-8 Algebraic Identities Exe-12 B, img 18

(vi)ICSE Class-8 Algebraic Identities Exe-12 B, img 19


Solved Questions of Exe-12 C for ICSE Class-8th Mathematics

Question 1 :-

If a  +b = 5 and ab = 6; find a2 + b2
Answer :-

(a + b)2 = a+ b2 + 2ab

⇒ (5)2 = a2 + b+ 2×6

⇒ 25 = a2 + b2 + 12

⇒ 25 − 12 = a2 + b2

⇒ 13 = a2 + b2

∴ a2 + b2 = 13

Question 2 :-

If a – b = 6 and ab = 16; find a2 + b2
Answer :-

(a − b)2 = a+ b2 − 2ab

⇒ (6)= a2 + b2 − 2×16

⇒ 36 = a2 + b2 − 32

⇒ 36 + 32 = a2 + b2

⇒ 68 = a2 + b2

∴ a2 + b2 = 68

Question 3 :-

If a2 + b2 = 29 and ab = 10 ; find :
(i) a + b
(ii) a – b
Answer :-

(i)(a + b)2 = a2 + b2  + 2ab

⇒ (a + b)2 = 29 + 2×10

⇒ (a + b)2 = 29 + 20

⇒ (a + b)2 = 49

⇒ a + b = √49

⇒ a + b = ±7

(ii)(a − b)2 = a2 + b2  − 2ab

⇒ (a − b)2 = 29 − 2×10

⇒ (a − b)2 = 29 − 20

⇒ (a − b)2 = 9

⇒ a − b = √9

⇒ a − b = ±3

Question 4 :-

If a2 + b= 10 and ab = 3; find :
(i) a – b
(ii) a + b
Answer :-

(i)

(a − b)2 = a2 + b2 − 2ab

⇒ (a − b)2 = 10 − 2×3

⇒ (a − b)2 = 10 − 6

⇒ (a − b)= 4

⇒ (a − b) = √4

⇒ a − b = ±2

(ii)

(a + b)2 = a2 + b2 + 2ab

⇒ (a + b)2 = 10 + 2×3

⇒ (a + b)2 = 10 + 6

⇒ (a + b)=  16

⇒ (a + b) = √16

⇒ a + b = ±4

Question 5 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-5
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 20

Question 6 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-6
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 21

Question 7 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-7
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 22

Question 8 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-8
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 23

Question 9 :-

If a + b + c = 10 and a2 + b2 + c2 = 38; find ab + bc + ca
Answer:-

a + b + c = 10

⇒ (a + b + c)= (10)2

⇒ a+ b2 + c+ 2ab + 2bc + 2ca = 100

⇒ 38 + 2(ab + bc + ca) = 100

⇒ 2(ab + bc + ca) = 100 − 38

⇒ 2(ab + bc + ca) = 62

⇒ (ab + bc+ ca) =62/2

⇒ ab + bc + ca = 31

Question 10 :-

Find a2 + b2 + c2 ; if a + b + c = 9 and ab + bc + ca = 24
Answer :-

a+ b + c = 9

⇒ (a + b + c)2 = (9)2

⇒ a2 + b+ c2 + 2ab + 2bc + 2ca = 81

⇒ a+ b2 + c+ 2(ab + bc + ca) = 81

⇒ a2 + b2 + c+ 2×24 = 81

⇒ a2 + b+ c2  + 48 = 81

⇒ a2 + b+ c= 81 − 48

⇒ a2 + b2 + c2 = 33

Question 11 :-

Find a + b + c; if a2 + b2 + c2 = 83 and ab + bc + ca = 71
Answer :-

(a + b + c)2 = a2 + b+ c+ 2ab + 2bc + 2ca

⇒ (a + b + c)2 = 82 + 2(ab + bc + ca)

⇒ (a + b+ c)2 = 83 + 2×71

⇒ (a + b + c)= 83 + 142

⇒ (a + b + c)= 225

⇒ a + b + c = √225

⇒ a + b +c = ±15

Question 12 :-

If a + b = 6 and ab=8; find a3 + b3
Answer :-

a + b = 6

⇒ (a + b)3 = (6)3

⇒ a+ b3 + 3ab (a + b) = 216

⇒ a3 + b+ 3×8 (6) = 216

⇒ a3 + b+ 144 = 216

⇒ a3 + b3 = 216 − 144

⇒ a3 + b3 = 72

Question 13 :-

If a – b = 3 and ab = 10; find a3 – b3
Answer :-

a − b = 3

⇒ (a − b)3 = (3)3

⇒ a3 − b3 − 3ab (a − b) = 27

⇒ a3 − b− 3×10 (3) = 27

⇒ a3 − b− 90 = 27

⇒ a3 − b3 = 27 + 90

⇒ a3 − b3 = 117

Question 14 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-14
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 24

Question 15 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-15
Answer :-

ICSE Class-8 Algebraic Identities Exe-12 B, img 25

Question 16 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-16
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 26

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 27

Question 17 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-C-17
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 28

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 29

Question 18 :-

The sum of the squares of two numbers is 13 and their product is 6. Find:
(i) the sum of the two numbers.
(ii) the difference between them.
Answer :-

Let x and y be the two numbers, then,

x2 + y2 = 13 and xy = 6

(i) (x + y)2 = x2 + y2 + 2xy

= 13 + 2×6

= 13 + 12

= 25

∴ x + y = ± √25 = ±5

(ii) (x − y)2 = x2 + y− 2xy

= 13 − 12

= 1

∴ x − y = ±1


Algebraic Identities ICSE Class-8th Maths Exe- 12 D

Question 1 :-

Evaluate:
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-1
Answer :-

(i) ICSE Class-8 Algebraic Identities Exe-12 B, img 30

(ii)

(2a + 0.5) (7a − 0.3)

= 2a (7a − 0.3) + 0.5 (7a − 0.3)

= 14a− 0.6a + 3.5a − 0.15

= 14a2 + 2.9a − 0.15

(iii)

(9 − y) (7 + y) = 9(7 + y) − y (7 + y)

= 63 + 9y − 7y − y2

= 63 + 2y − y2

(iv)

(2 − z) (15 − z) = 2(15 − z) −z(15 − z)

= 30 − 2z − 15z + z2

= 30 − 17z + z2

(v)

(a2 + 5) (a2 − 3) = a2 (a2 − 3) + 5 (a2 − 3)

= a− 3a2 + 5a2 − 15

= a4 + 2a2 − 15

(vi)

(4 − ab) (8 + ab) = 4 (8 + ab) − ab (8 + ab)

= 32 + 4ab − 8ab − a2b2

= 32 − 4ab − a2b2

(vii)

(5xy − 7) (7xy + 9) = 5xy (7xy + 9) − 7 (7xy + 9)

= 35x2y+ 45xy − 49xy − 63

= 35x2y2 − 4xy − 63

(viii)

(3a2 − 4b2) (8a2 − 3b2)

= 3a(8a2 − 3b2) − 4b2 (8a2 − 3b2)

= 24a4 − 9a2b2 − 32a2b2 + 12b4

= 24a−  41a2b2 + 12b4

Question 2 :-

Evaluate:
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-2
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 31

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 32

(iii)

(6 − 5xy) (6 + 5xy)

= (6)− (5xy)2

= 36 − 25x2y2   ………[ ∵ (a − b) (a + b) = a2b2]

(iv)ICSE Class-8 Algebraic Identities Exe-12 B, img 33

(v)

(4x2 − 5y2) (4x2 + 5y2)

= (4x)2 − (5y2)

= 16x4 − 25y4 ……[ ∵ (a − b) (a + b) = a2 − b2]

(vi)

(1.6x + 0.7y) (1.6x − 0.7y)

= (1.6x)2 − (0.7y)……[ ∵ (a − b) (a +b) = a2 − b2]

= 2.56x2 − 0.49y2

(vii)

(m + 3) (m − 3) (m2 + 9)

= (m)− (3)2 (m2 + 9) ……[ ∵ (a − b) (a + b) = a2 − b2]

= (m− 9) (m2 + 9)

= (m2)− 92

= m− 81

(viii)

(3x + 4y) (3x − 4y) (9x2 + 16y2)

= [(3x)2 − (4y)2] (9x2 + 16y2) …..[ ∵ (a − b) (a + b) = a2 − b2]

= (9x2 − 16y2) (9x2 + 16y2)

= (9x2)2 − (16y2)2  ……[ ∵ (a − b) (a + b) = a− b2]

= 81x4 − 256y4

(ix)

(a + bc) (a − bc) (a2 + b2c2)

= [a2 − (bc)2] (a2 + b2c2)  …….[ ∵ (a − b) (a + b) = a2 − b2]

= (a2 − b2c2) (a+ b2c2)

= (a2)2 − (b2c2)2 …….[ ∵ (a − b) (a + b) = a2 − b2]

= a4 − b4c4

(x)

203 × 197

= (200 + 3) (200 − 3)

= (200)− (3)2  ……[ ∵ (a − b) (a + b) = a2 − b2]

= 40000 − 9

= 39991

(xi)

20.8 × 19.2

= (20 + 0.8) (20 − 0.8)

= (20)− (0.8)2  ……[ ∵ (a − b) (a + b) = a2 − b2]

= 400 − 0.64

= 399.36

Question 3 :-

Find the square of :
selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-3
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 34

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 35

(iii)ICSE Class-8 Algebraic Identities Exe-12 B, img 36

(iv)ICSE Class-8 Algebraic Identities Exe-12 B, img 37

(v)ICSE Class-8 Algebraic Identities Exe-12 B, img 38

(vi)

(607)2 = (600 + 7)2

= (600)2 + (7)2 + 2 (600) (7)

= 360000 + 49 + 8900

= 368449

(vii)

(319)2 = (400 − 9)2

= (400)2 + 92 + 2 (400) (9)

= 160000 + 81 − 7200

= 152881

(vii)

(9.7)2 = (10 − 0.3)2

= (10)2 + (0.3)2 − 2 (10) (0.3)

= 100 + 0.9 − 6

= 100.09 − 6.00

= 94.09

Question 4 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-4
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 39

= 4 − 2

= 2

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 40

= 4 − 2

= 2

Question 5 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-5
Answer :-

(i)ICSE Class-8 Algebraic Identities Exe-12 B, img 41

= (5)2 + 2

= 25 + 2

= 27

(ii)ICSE Class-8 Algebraic Identities Exe-12 B, img 42

= (27)2 − 2

= 729 − 2

= 727

(iii)ICSE Class-8 Algebraic Identities Exe-12 B, img 43

Question 6 :-

If a2 + b2 = 41 and ab = 4, find :
(i) a – b
(ii) a + b
Answer :-

(i)

(a − b)2 = a2 + b2 − 2ab

= 41 − 2(4)

= 41 −  8

= 33

∴ a − b = √33

(ii)

(a + b)2 = a2 + b2 + 2ab

= 41 + 2(4)

= 41 +  8

= 49

⇒ (a + b)2 = 49

∴ a + b = 7

Question 7 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-7
Answer :-

(i)

ICSE Class-8 Algebraic Identities Exe-12 B, img 44

= (8)2 − 2

= 64 − 2

= 62

(ii)

ICSE Class-8 Algebraic Identities Exe-12 B, img 45

= (62)2 − 2

= 3844 − 2

= 3842

Question 8 :-

selina-concise-mathematics-class-8-icse-solutions-algebraic-identities-D-8
Answer :-

(i)

ICSE Class-8 Algebraic Identities Exe-12 B, img 46

= (5)2 + 2

= 25 + 2

= 27

(ii)

ICSE Class-8 Algebraic Identities Exe-12 B, img 47

= (27)2 − 2

= 729 − 2

= 727

Question 9 :-

Expand :
(i) (3x – 4y + 5z)2
(ii) (2a – 5b – 4c)2
(iii) (5x + 3y)3
(iv) (6a – 7b)3
Answer :-

(i)

(3x – 4y + 5z)2

= (3x)+ (– 4y)2 + (5z)+ 2(3x) (– 4y) + 2(– 4y) (5z) + 2(5z) (3x)

= 9x2 + 16y2 + 25z2 – 24xy – 40yz + 30zx

(ii)

(2a – 5b – 4c)2

= (2a)2 + (–5b)2 + (–4c)2 + 2 (2a) (–5b) + 2 (–5b) (–4c) + 2 (–4c) (2a)

= 4a2 + 25b2 + 16c2 – 20ab + 40bc – 16ca

(iii)

(5x + 3y)3

= (5x)3 + (3y)3 + 3 (5x) (3y) (5x + 3y)

= 125x3 + 27y3 + 45xy (5x + 3y)

= 125x+ 27y+ 225x2y + 135xy2

(iv)

(6a – 7b)3

= (6a)3 – (7b)3 – 3(6a) (7b) (6a – 7b)

= 216a3 – 343b– 126ab (6a – 7b)

= 216a3 – 343b3 – 756a2b + 882ab2

= 216a3 – 756a2b + 882ab2 – 343b3

Question 10 :-

If a + b + c = 9 and ab + bc + ca = 15, find: a2 + b2 + c2.
Answer :-

Since (a + b + c)= a2 + b2 + c2 + 2 (ab + bc + ca)

∴ (9)2 = a2 + b+ c2 + 2 (15)

81 = a+ b2 + c2 + 30

∴ a2 + b2 + c= 81 − 30 = 51

Question 11 :-

If a + b + c = 11 and a2 + b2 + c2 = 81, find ab + bc + ca.
Answer :-

Since (a + b + c)2 = a+ b2 + c2 + 2 (ab + bc + ca)

∴ (11)2 = 81 + 2 (ab + bc + ca)

∴ 2(ab + bc + ca) = 121 − 81 = 40

ab + bc + ca = 40/2

⇒ ab + bc + ca = 20

Question 12 :-

If 3x – 4y = 5  and xy = 3, find : 27x3 – 64y3.
Answer :-

27x3 – 64x3 = (3x)3 – (4y)3

= (3x − 4y)3 (3x − 4y)3 + 3 (3x) (4y) (3x − 4y) ……[ ∵ a3 − b= (a − b)3 + 3ab(a − b)]

= (5)3 + 36(xy) (3x − 4y)

= 125 + 36 (3) (5)

= 125 + 540

= 665

Question 13 :-

If a + b = 8 and ab = 15, find : a3 + b3.
Answer :-

a3 + b3 = (a + b)3 − 3ab (a + b)

= (8)3 − 3(15) (8)

= 512 − 360

= 152

Question 14 :-

If 3x + 2y = 9 and xy = 3, find : 27x3 + 8y3
Answer :-

27x3 + 8y3 = (3x)3 + (2y)3

= (3x + 2y)3 − 3.3x . 2y (3x + 2y)

= (3x − 2y)3 − 18xy (3x + 2y)

= (9)− 18(3) (9)

= 729 − 486

= 243

Question 15 :-

If 5x – 4y = 7 and xy = 8, find : 125x3 – 64y3
Answer :-

125x3 − 64y3 = (5x)3 − (4y)3

= (5x − 4y)3 + 3(5x) (4y) (5x − 4y)

= (5x − 4y)3 + 60xy (5x − 4y)

= (7)+ 60 (8) (7)

= 343 + 3360

= 3703

Question 16 :-

The difference between two numbers is 5 and their products is 14. Find the difference between their cubes.
Answer :-

Let x and y be two numbers, then x – y = 5 and xy = 14

∴ x3 − y3 = (x − y)3 + 3xy(x − y)

= (5)3 + 3 × 14 × 5

= 125 + 210

= 335

 

— End of Algebraic Identities Solutions :–

Return to Concise Selina Maths Solutions for ICSE Class -8 

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