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
a2 − b2 we have a = 2p and b
= 5q
(a2 − b2) = (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)3
= (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
+ 3ab2 − b3
(5 − x)3 = 53 − 3 (5)2
(x) + 3(5)(x2) − x3
= 125 − 3(25)(x) + 15x2 − x3
= 125 − 75x + 15 x2 − x3
(ii) (2x − 4y)3
Comparing (2x − 4y)3 with (a − b)3
we have a = 2x and b = 4y
(a − b)3 = a3 − 3a2b
+ 3ab2 − b3
(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
+ 3ab2 − b3
(ab − c)3 = (ab)3 − 3(ab)2
c + 3ab (c)2 − c3
= a3b3 − 3(a2b2)
c + 3abc2 − c3
= a3b3 − 3a2b2
c + 3abc2 − c3
(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
+ 3ab2 − b3
(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
+ 3ab2 − b3
(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 x2–y2
= 16 and (x+y) = 8 then (x–y) is ___________
(A) 8
(B) 3
(C) 2
(D) 1
[Answer: (C) 2]
Solution:
x2− y2 = 16
(x − y) (x − y) = 16
8 (x − y) = 16
(x − y) = 16 / 8 = 2
8.
(A) a2–ab+b2
(B) a2+ab+b2
(C) a2+2ab+b2
(D) a2–2ab+b2
[Answer: (B) a2 + ab + b2 ]
Solution:
[(a − b)2 (a3 – b3)
] / [(a2 – b2)] = [ (a + b) (a – b)
a2 + ab + b2 )] / [ (a + b)
(a – b)]
= a2 + ab
+ b2
9. (p+q)(p2–pq+q2) is equal to _____________
(A) p3+q3
(B) (p+q)3
(C) p3–q3
(D) (p–q)3
[Answer: (A) p3 + q3 ]
Solution:
a3 + b3 =
(a + b)(a2 − ab + b2)
10. (a–b)=3 and ab=5 then a3–b3
= ____________
(A) 15
(B) 18
(C) 62
(D) 72
[Answer: (D) 72]
Solution:
(a − b) = 3
(a − b)2 = 32
a2 + b2 − 2ab = 9
a2 + b2 – 2(5)
= 9
a2 + b2 = 9 + 10
a2 + b2 = 19
a3 + b3 = (a – b) (a2
+ ab + b2) = 3 (19 + 5)
= 3(24) = 72
11. a3+b3 = (a+b)3 – __________
(A) 3a(a+b)
(B) 3ab(a–b)
(C) –3ab(a+b)
(D) 3ab(a+b)
[Answer: 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
Answer
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 + 3abc3 − c3
(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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