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# Exercise 3.3 (Cubic Identities)

8th Maths : Chapter 3 : Algebra : Cubic Identities : Exercise 3.3: Text Book Back Exercises Questions with Answers, Solution

Exercise 3.3

1. Expand

(i) (3m + 5)2

(ii) (5p 1)2

(iii) (2n 1)(2n + 3)

(iv) 4 p 2 25q2

Solution:

(i) (3m + 5)2

Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5

(a + b)2 = a2 + 2ab + b2

(3m + 5)2 = (3m)2 + 2 (3m) (5) + 52

= 32 m2 + 30 m + 25 = 9m2 + 30m + 25

(ii) (5p − 1)2

Comparing (5p − l)2 with (a − b)2 we have a = 5p and b = 1

(a − b)2 = a2 − 2ab + b2

(5p − l)2 = (5p)2 − 2 (5p) (l) + l2

= 52p2 − 10p + 1 = 25p2 − 10p + 1

(iii) (2n − 1) (2n + 3)

Comparing (2n − 1) (2n + 3) with (x + a) (x + b) we have a = − l; b = 3

(x + a) (x + b)  = x2 + (a + b)x + ab

(2n + (− 1)) (2n + 3) = (2n)2 + (−1 + 3) 2n + (−1) (3)

= 22n2 + 2 (2n) − 3= 4n2 + 4n − 3

(iv) 4p2 − 25q2 = (2p)2 − (5q)2

Comparing (2p)2 − (5q)2 with a2b2 we have a = 2p and b = 5q

(a2b2) = (a + b) (a − b) = (2p + 5q) (2p − 5q)

2. Expand

(i) (3 + m)3

(ii) (2a + 5)3

(iii) (3 p + 4q)3

(iv) (52)3

(v) (104)3

Solution:

(i) (3 + m)3

Comparing (3 + m)3 with (a + b)3 we have a = 3 ;  b = m

(a + b)3 = a2 + 3a2b + 3ab2 + b3

(3 + m)3 = 33 + 3(3)2(m) + 3(3) m2 + m3

= 27 + 27m + 9m2 + m3 = m3 + 9m2 + 27m +  27

(ii) (2a + 5)3

Comparing (2a + 5)3 with (a + b)3 we have a = 2a, b = 5

(a + b)3 = a3 + 3a2b + 3ab2 + b3

= (2a)3 + 3(2a)2 5 + 3 (2a) 52 + 53

= 23a3 + 3(22a2) 5 + 6a (25) + 125

= 8 a3 + 60 a2 + 150a + 125

(iii) (3p + 4q)3

Comparing (3p + 4q)3 with (a + b)3 we have a = 3p and b = 4q

(a + b) = a3 + 3a2b + 3ab2 + b2

(3p + 4q)= (3p)3+ 3(3p)2 (4q) + 3(3p) (4q)2 + (4q)3

= 33p3 + 3 (9p2) (4q) + 9p (16q2) + 43q3

= 27p3 + 108p2q + 144pq2 + 64q3

(iv) (52)3 = (50+ 2)3

Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b = 2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(50 + 2)3 = 503 + 3 (50)2 2 + 3 (50)(2)2 + 23

523 = 125000 + 6(2,500) + 150(4) + 8

= 1,25,000 + 15,000 + 600 + 8

523 = 1,40,608

(v) (104)3 = (100+ 4)3

Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + 43

= 10,00,000 + 3 (10000) 4 + 300 (16) + 64

= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

3. Expand

(i) (5 x)3

(ii) (2x 4 y)3

(iii) (ab c)3

(iv) (48)3

(v) (97xy)3

Solution:

(i) (5 − x)3

Comparing (5 − x)3 with (a − b)3 we have a = 5 and b = x

(a − b)3 = a3 − 3a2b + 3ab2b3

(5 − x)3 = 53 − 3 (5)2 (x) + 3(5)(x2) − x3

= 125 − 3(25)(x) + 15x2x3 = 125 − 75x + 15 x2x3

(ii) (2x − 4y)3

Comparing (2x − 4y)3 with (a − b)3 we have a = 2x and b = 4y

(a − b)3 = a3 − 3a2b + 3ab2b3

(2x − 4y)3 = (2x)3 − 3(2x)2 (4y) + 3(2x) (4y)2 − (4y)3

= 23 x3 − 3(22 x2) (4y) + 3(2x) (42 y2) − (43y3)

= 8x3 − 48x2y + 96xy2 − 64y3

(iii) (ab − c)3

Comparing (ab − c)3 with (a − b)3 we have a = ab and b = c

(a − b)3 = a3 − 3a2b + 3ab2b3

(ab − c)3 = (ab)3 − 3(ab)2 c + 3ab (c)2c3

= a3b3 − 3(a2b2) c + 3abc2c3

= a3b3 − 3a2b2 c + 3abc2c3

(iv) (48)3 = (50 −2)3

Comparing (50 − 2)3 with (a − b)3 we have a = 50 and b = 2

(a − b)3 = a3 − 3a2b + 3ab2b3

(50 − 2)3 = (50)3 − 3(50)2 (2) + 3 (50)(2)2 − 23

= 1,25,000 − 15000 + 600 − 8 = 1,10,000 + 592

= 1,10,592

(v) (97xy)3

(97xy)3 = 973 x3y3 = (100 − 3)3 x3y3 ... (1)

Comparing (100 − 3)3 with (a − b)3 we have a = 100, b = 3

(a – b)3 = a3 − 3a2b + 3ab2b3

(100 − 3)3 = (100)3 − 3(100)2 (3) + 3 (100)(3)2 − 33

973 = 10,00,000 − 90000 + 2700 − 27

973 = 910000 + 2673

973 = 912673

97x3y3 = 912673x3y3

4. Simplify ( p 2)( p + 1)( p 4)

Solution:

(p − 2)(p + 1) (p − 4) = (p + (−2)) (p + l) (p + (−4))

Comparing (p − 2) (p + 1) (p − 4) with (x + a) (x + b) (x + c) we have x = p ; a = −2; b = 1 ; c = −4.

(x + a) (x + b) (x + c) = x2 + (a + b + c) x2 + (ab + bc + ca) x + abc

= p3 + (−2 + 1 + (−4)) p2 + ((−2) (1) + (1) (−4) + (−4) (−2))p + (−2) (1) (−4)

= p3 + (−5) p2 + (−2 + (−4) + 8) p + 8

= p3 − 5p2 + 2p + 8

5. Find the volume of the cube whose side is (x +1) cm

Solution:

Given side of the cube = (x + 1) cm

Volume of the cube = (side)3 cubic units = (x + 1)3 cm3

We have (a + b)3 = (a3 + 3a2b + 3ab2 + b3) cm3

(x + 1)3 = (x3 + 3x2 (1) + 3x (1)2 + l3) cm3

Volume = (x3 + 3x2 + 3x + 1) cm3

6. Find the volume of the cuboid whose dimensions are (x + 2),(x 1) and (x 3)

Solution:

Given the dimensions of the cuboid as  = (x + 2), (x – 1) and (x – 3)

Volume of the cuboid = (l × b × h) units3

= (x + 2) (x − 1) (x − 3) units3

We have (x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc+ ca)x + abc

(x + 2) (x − l) (x − 3) = x3 + (2 − 1 − 3)x2 + (2 (−1) + (−1) (−3) + (−3) (2)) x +(2)(−l) (−3)

= x3 – 2x2 + (−2 + 3 − 6)x + 6

Volume = x3 – 2x2 − 5x + 6 units3

Objective Type Questions

7. If x2y2 = 16 and (x+y) = 8 then (xy) is ___________

(A) 8

(B) 3

(C) 2

(D) 1

Solution:

x2y2 = 16

(x − y) (xy) = 16

8 (xy) = 16

(xy) = 16 / 8 = 2

8.

(A) a2ab+b2

(B) a2+ab+b2

(C) a2+2ab+b2

(D) a2–2ab+b2

[Answer: (B) a2 + ab + b2 ]

Solution:

[(ab)2 (a3b3) ] / [(a2b2)] =  [ (a + b) (ab) a2 + ab + b2 )] / [ (a + b) (ab)]

= a2 + ab + b2

9. (p+q)(p2pq+q2) is equal to _____________

(A) p3+q3

(B) (p+q)3

(C) p3q3

(D) (pq)3

[Answer: (A) p3 + q3 ]

Solution:

a3 + b= (a + b)(a2ab + b2)

10. (ab)=3 and ab=5 then a3b3 = ____________

(A) 15

(B) 18

(C) 62

(D) 72

Solution:

(ab) = 3

(ab)2 = 32

a2 + b2 − 2ab  = 9

a2 + b2 – 2(5)  = 9

a2 + b2 = 9 + 10

a2 + b2 = 19

a3 + b3 = (ab) (a2 + ab + b2) = 3 (19 + 5)

= 3(24) = 72

11. a3+b3 = (a+b)3 – __________

(A) 3a(a+b)

(B) 3ab(ab)

(C) –3ab(a+b)

(D) 3ab(a+b)

Solution:

(a + b)3 = a3 + b3 + 3a2b + 3ab2

(a + b)3 – 3a2b − 3ab2 = a3 + b3

(a + b)3 – 3ab (a + b)= a3 + b3

Exercise 3.3

1. (i) 9m2 + 30 m + 25  (ii) 25 p2 − 10p +1  (iii ) 4n2 + 4n − 3 (iv) ( 2p + 5q )( 2p − 5q)

2. (i) m3 + 9m2 + 27 m + 27 (ii) 8a3 + 60 a2 +150 a +125 (iii ) 27 p3 + 108 p2q +144 pq2 + 64q3 (iv) 14,0608 (v) 1124864

3. (i) 125 − 75 x +15x2 x3 (ii) 8x3 − 48x2 y + 96 xy2 − 64y3 (iii ) a3b3 − 3a2b2c + 3abc3c3

(iv) 110, 592 ( v ) 912673x3y3

4. p3 − 5p2 + 2p + 8

5. x3 + 3x2 + 3 x +1

6. x3 − 2x2 − 5 x + 6

7. (C) 2

8. (B) a2 + ab + b2

9. (A) p3 + q3

10. (D) 72

11. (D) 3ab(a+b)

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8th Maths : Chapter 3 : Algebra : Exercise 3.3 (Cubic Identities) | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths