ML Aggarwal Solutions Class 9 Math 3rd Chapter Expansions Exercise 3.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Third Chapter Expansions Exercise 3.1. APC Solution Class 9 Exercise 3.1.
(1) (i) (2x + 7y)2
Solution:
(2x + 7y)2
= (2y)2 + 2 (2x) (7y) + (7y)2
= 4x2 + 28xy + 49y2
(ii) (1/2 x + 2/3 y)2
Solution:
(1/2 x + 2/3 y)2
= (1/2 x)2 + 2 (1/2 x) (2/3 y) + (2/3 y)2
= x2/4 + 2xy/3 + 482/9
(2) (i) (3x + 1/2x)2
Solution:
(3x + 1/2x)2
= (3x)2 + 2 (3x) (1/2x) + (1/2x)2
= 9x2 + 3 + 1/4x2
(ii) (3x2y + 5z)2
Solution:
= (3x2y)2 + 2 (3x2y) (57) + (57)
= 9x4 y2 + 36x2yz + 2572
(3) (i) (3x – 1/2x)2
Solution:
= (3x)2 – 2x (3x) (1/2x) + (1/2x)2
= 9x2 – 3 + 1/4x2
(ii) (1/2 x – 3/2 y)2
Solution:
(1/2 x)2 – 2x (1/2y x) (3/2 y) + (3/2 y)2
= x2/4 – 3yx/2 + 9y2/4
(4) (i) (x + 3) (x + 5)
Solution:
= x2 + (3+5) x + 3×5
= x2 + 8x + 15
(ii) (x + 3) (x – 5)
Solution:
= x2 – 5x + 3x – 3 × 5
= x2 – 2x – 15
(iii) (x – 7) (x + 9)
Solution:
= x2 + 9x – 7x – 7 × 9
= x2 + 2x – 63
(iv) (x – 2y) (x – 3y)
Solution:
= x2 – (2y + 3y) x + 2y × 3y
= x2 – 5xy + 6y2
(5) (i) (x – 2y – z)2
Solution:
= (x – 2y – 7) (x – 2y – 7)
= x2 – 2xy – x7 – 2xy + 4y2 + 2yz – xz + 2yz + z2
= x2 + 4y2 + z2 – 4xy – 2xz + 4yz
(ii) (2x – 3y + 4z)2
Solution:
= (2x – 3y + 4z) (2x – 3y + 4z)
= 4x2 – 6xy + 8xz – 6xy + 9y2 – 12yz + 8xz – 12yz + 18z2
= 4x2 + 9y2 + 16z2 – 12xy + 16xz – 24yz
(6) (i) (2x + 3/x – 1)2
Solution:
= 4x2 + 2x.3/x – 2x + 3/x.2x + 9/x2 – 8/x – 2x – 3/x + 1
= 4x2 + 9/x2 + 1 – 4x + 12 – 6/x
= 4x2 + 9/x2 – 6/x – 4x + 13
(ii) (2/3 x – 3/2x – 1)2
Solution:
= (2/3 x – 3/2x – 1) (2/3 x – 3/2x – 1)
= 4/4 x2 – 2/3 x. 3/2x – 2/3 x – 3/2x.2/3 x + 9/4 x2+ 3/2x – 2/3 x + 3/2x + 1
= 4/9 x2 + 9/4x – 4/3 x + 6/2x – 2+1
= 4/9 x2 + 9/4x2 – 4x/3 + 6/2x – 1
(7) (i) (x + 2)3
Solution:
= x3 + 23 + 3x × 2 (x + 2)
= x3 + 8 + 6x2 + 12x
= x3 + 6x2 + 12x + 8
(ii) (2a + b)3
Solution:
= (2a)3+ b3 + 3.2a.b (2a + b)
= 8a3 + b3 + 12a2b + ab2
(8) (i) (3x + 1/x)3
Solution:
= (3x)3 + 1/x3 + 3 × 3x.1/x (3x + 1/x)
= 27x3 + 1/x3 + 27x + 9/x
(ii) (2x – 1)3
Solution:
= (2x)3 – 13 – 3.2x.1 (2x – 1)
= 8x3 – 1 – 12x2 + 6x
(9) (i) (5x – 3y)3
Solution:
= (5x)3 – (3y)3 – 3.5x.3y (5x – 3y)
= 125x3 – 27y3 – 225x2y + 135xy2
(ii) (2x – 1/3y)3
Solution:
= (2x)2 – (1/3y)3 – 3.2x.1/3y (2x – 1/3y)
= 8x3 – 1/27y3 – 4x2/y + 2x/3y2
Simplify the following (10 to 19):
(10) (i) (a + b)2 + (a – b)2
Solution:
= a2 + 2ab + b2 + a2 – 2ab + b2
= 2a2 + 2b2
= 2 (a2 + b2)
(ii) (a + b)2 – (a – b)2
Solution:
(a + b)2 – (a – b)2
= a2 + 2ab + b2 – a2 + 2ab – b2
= 4ab
(11) (i) (a + 1/a)2 + (a – 1/a)2
Solution:
= a2 + 1/a2 + 2 + a2 + 1/a2 – 2
= 2a2 + 2/a2
= 2 (a2 + 2/a2)
(ii) (a + 1/a)2 – (a – 1/a)2
Solution:
= a2 + 1/a + 2 – a2 – 1/a2 + 2
= 4
(12) (i) (3x – 1)2 – (3x – 2) (3x + 1)
Solution:
= 9x2 – 2.3x.1 + 12 – (9x2 + 3x – 6x – 2)
= 9x2 – 6x + 1 – 9x2 + 3x + 2
= 3 – 3x
= 3 (1 – x)
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy
Solution:
= 16x2 + 2.4x.3y+ 9y2 – (46x2 – 2.4x.3y + 9y2) – 48xy
= 16x2 + 24xy + 9y2 – 16x2 + 24xy – 9y2 – 48xy
= 0
(13) (i) (7p + 9q) (7p – 9q)
Solution:
= (7p)2 – (9q)2
= 49q2 – 81q2
(ii) (2x – 3/x) (2x + 3/x)
Solution:
= (2x)2 – (3/x)2
= 4x2 – 9/x2
(14) (i) (2x – y + 3) (2x – y – 3)
Solution:
= 4x2 – 2xy – 6x – 2xy + y2 + 3y + 8x – 3y – 9
= 4x2 + y2 – 4xy – 9
(ii) (3x + y – 5) (3x – y – 5)
Solution:
= (3x – 5 + y) (3x – 5 – y)
= (3x – 5)2 – y2
= 9x2 – 2.3x.5 + 52 – y2
= 9x2 – y2 – 30x + 25
(15) (i) (x + 2/x – 3) (x – 2/x – 3)
Solution:
= (x – 3 + 2/) (x – 3 – 2/x)
= (x – 3)2 – (2/x)2
= x2 – 2.3.x + 9 – 4/x2
= x2 – 6x + 9 – 4/x2
(ii) (5 – 2x) (5 + 2x) (25 + 4x2)
Solution:
= {52 – (2x)2} (25 + 4x2)
= (25 – 4x2) (25 + 4x2)
= 252 – (4x2)2
= 625 – 16x4
(16) (i) (x + 2y + 3) (x + 2y + 7)
Solution:
= x2 + 2xy + 7x + 2xy + 4y2 + 7.14y+ 3x + 6y + 21
= x2 + 4y2 + 4xy + 18x + 20y + 21
(ii) (2x + y + 5) (2x + y – 9)
Solution:
= 4x2 + 2xy – 18x + 2xy + y2 – 9y + 10x + 5y – 45
= 4x2 + y2 + 4xy – 8x – 4y – 45
(iii) (x – 2y – 5) (x – 2y + 3)
Solution:
= x2 – 2xy + 3x – 2xy – 4y2 – 6y – 5x + 10y – 15
= x2 – 4y2 – 4xy – 2x + 4y – 15
(iv) (3x – 4y – 2) (3x – 4y – 6)
Solution:
= 9x2 – 12xy – 18x – 12xy + 16y2 + 24y – 6x + 8y + 12
= 9x2 + 16y2 – 24xy – 24x + 32y + 12
(17) (i) (2p + 3q) (4p2 – 6pq + 9q2)
Solution:
= (2p + 3q) {(2p)2– 2p.3q + (3q)2}
= (2p)3 + (3q)3
= 8p3 + 27q3
(ii) (x + 1/x) (x2 – 1 + 1/x2)
Solution:
= (x + 1/x) {(x)2– x. 1/x + (1/x)2}
= x3 + 1/x3
(18) (i) (3p – 4q) (9p2 + 12pq + 16q2)
Solution:
= (3p – 4q) {(3p)2 + 3p.4q + (4q)2}
= (3p)3 – (4q)3
= 27p3 – 64q3
(ii) (x – 3/x) (x2 + 3 + 9/x2)
Solution:
= (x – 3/x) {(x)2 + x .3/x + (3/x)2}
= x3 – (3/x)3
= x3 – 27/x3
(19)(2x + 3y + 4z) (4x2 + 9y2 + 16z2 – 6xy – 12yz – 8zx)
Solution:
= (2x + 3y + 4z) {(2x)2 + (3y)2 + (4z)2 – 2x.3y – 3y.4z – 2x.4z}
= (2x)3 + (3y)3 + (4z)3 – 3.2x.3y.4z
= 8x3 + 27y3 + 64z3 – 72xyz
(20) (i) Find the product of the following:
(i) (x + 1) (x + 2) (x + 3)
Solution:
= x3 + (1 + 2 + 3) x2 + (1 × 2 + 2 × 3 + 1 × 3) x + 1 × 2 × 3
= x3 + 6x2 + 11x + 6
(ii) (x – 2) (x – 3) (x + 4)
Solution:
= x3+ (-2 – 3 + 4) x2 + (-2 × -3 + (-3 × 4) + (-2 × 4)) + -2 × -3 × 4
= x3 – x2 – 14x + 24
(21) Find the coefficient of x2 and x in the product of (x – 3) (x + 7) (x – 4)
Solution:
(x – 3) (x + 7) (x – 4)
= x3 + (-3 + 7 – 4) x2 + (-3 × 7 + (7 × -4) + (-3 × -4)) x + -3 × 7 x – 4
= x3 + 0.x– 37x + 84
∴ Coefficient of x2 is O, coefficient of x is -37
(22) If a2 + 4a + x = (a + 2)2, find the value of x.
Solution:
Given, a2 + 4a + x = (a + 2)2
Or, a2 + 4a + x = a2 + 2.a.2 + 4
Or, a2 + 4a + x = a2 + 4a + 4
Or, x = 4
(23) Use (a + b)2 = a2 + 2ab + b2 to evaluate the following:
(i) (101)2
(ii) (1003)2
(iii) (10.2)2
Solution:
(i) (101)2
= (100 + 1)2
= 1002 + 2 × 100 × 1 + 12
= 10000 + 200 + 1
= 10201
(ii) (1003)2
= (1000+3)2
= 10002 + 2 × 1000 × 3 + 32
= 1000000 + 6000 + 9
= 1006009
(iii) (10.2)2
= (10+0.2)2
= 102 + 2 × 10 × 0.2 + (0.2)2
= 100 + 4 + 0.4
= 104.04
(24) Use (a – b)2 = a2 – 2ab + b2 to evaluate the following:
(i) (99)2
Solution:
= (100 – 1)2
= 1002 – 2 × 100 × 1 + 12
= 10000 – 200 + 1
= 9801
(ii) (997)2
Solution:
= (1000 – 3)2
= 10002 – 2 × 1000 × 3 + 32
= 1000000 – 6000 + 9
= 994009
(iii) (9.8)2
Solution:
= (10 – 0.2)2
= 102 – 2 × 10 × 0.2 + (0.2)2
= 100 – 4 + 0.04
= 96.04
(25) By using suitable identities, evaluate the following:
(i) (103)3
Solution:
= (100 + 3)3
= 1003 + 33 + 3 × 100 × 3 (100 + 3)
= 1000000 + 27 + 3 × 100 × 3 × 103
= 1000000 + 27 + 92700
= 1092727
(ii) (99)3
Solution:
= (100 – 1)3
= 1003 – 13 – 3 × 100 × 1 (100 – 1)
= 100000 – 1 – 300 × 99
= 1000000 – 1 – 29700
= 970299
(iii) (10.1)3
= (10 + 0.1)3
= 103 + (0.1)3 + 3 × 10 × 0.1 (10 + 0.1)
= 1000 + 0.001 + 3 × 10.01
= 1000 + 0.001 + 30.3
= 1030.301
(26) If 2a – b + c = 0, prove that 4a2 – b2 + c2 + 4ac = 0
Hint: 2a – b + c = 0 => (2a + c)2= b2
Solution:
Given, 2a – b + c = 0
Or, 2a + c = b
Or, (2a + c)2 = b2
Or, 4a2 + 2.2a.c + c2 = b2
Or, 4a2 + 4ac + c2 = b2
Or, 4a2 – b2 + c2 + 4ac = 0 (Proved)
(27) If a + b + 2c = 0, Prove that a3 + b3 + 8c3 = 6abc
Solution:
a + b + 2c = 0
Or, a + b = -2c — (i)
Or, (a + b)3 = (-2c)3
Or, a3 + b3 + 3ab (a + b) = -8c3
Or, a3 + b3 + 3a2b + 3ab2 = – 8c3
Or, a3 + b3 + 8c3 = -3a2 b + 3ab2
Or, a3 + b3 + 8c3 = – 3ab (a + b)
Or, a3 + b3 + 8c3 = – 3abx – 2c [From equation (i) a + b = 2c]
Or, a3 + b3 + 8c3 = 6abc (Proved)
(28) If a + b + c = 0, then find the value of a2/bc + b2/ca + c2/ab
Solution:
Given, a + b + c = 0
Or, a + b = -c —- (i)
Or, (a + b)3 = (-c)3
Or, a3 + b3 + 3ab (a + b) = – c3
Or, a3 + b3 + c3 = -3ab (a + b)
Or, a3 + b3 + c3 = -3ab × – c [From equation (i)]
Or, a3 + b3 + c3 = 3abc
Dividing both sides by abc we get,
Or, a3/abc + b3/abc + c3/abc = 2abc/abc
Or, a2/bc + b2/ac + c2/ab = 3
(29) If x + y = 4, then find the value of x3 + y3 + 12xy – 64.
Solution:
Given, x + y = 4 —- (i)
Or, (x + y)3 = 43
Or, x3 + y3 + 3xy (x+y) = 64
Or, x3 + y3 + 3xy × 4 = 64 [From — (i)]
Or, x3 + y3 + 12xy = 64
Or, x3 + y3 + 12xy – 64 = 0 (Ans)
(30) Without actually calculating the cubes, find the values of:
(i) (27)3 + (-17)3 + (-10)3
Solution:
(27)3 + (-17)3 + (-10)3
We know,
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – ac – bc)
Or, a3 + b3 + c3 = 3abc + (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
Putting the values, a = 27, b = -17, c = -10
(27)3 + (-17)3 + (-10)3 = 3 × 27 × – 17x – 10 + (27 – 17 – 10)
(272 + (-17)2 + (-10)2 – (27x – 17) – (-17x – 10) – (27x – 10))
= 13770 + 0 × (272 + 172 + 102 – (27x – 17) – (-17x – 10) – 27x – 10)
= 13770 + 0
= 13770
(ii) (-28)3 + (15)3 + (13)3
Solution:
We know,
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
Or, a3 + b3 + c3 = 3abc + (a + b + c) (a2 + b2 + c2 – ab – bc – ac)
Putting the value a = -28, b = 15, c = 13 we get,
RHS
3 × – 28 × 15 × 13 + (-28 + 15 + 13) (282 + 152 + 132 – (-28 × 15) – 15 × 13 – (-28 × 13))
= -16380 + 0 (283 + 152 + 13 – (-28 × 15) – 15 × 13 (-28 × 13))
= -16380 + 0
= – 16380
(31) Using suitable identity, find the value of:
86×86×86+14×14×14/86×86-86×14+14×14
Solution:
We know,
a3 + b3 = (a + b) (a2 – ab + b2)
Or, a3+b3/a2-ab+b2 = (a + b)
Putting the value of (a + b) on RHS,
Or, a3+b3/a2-ab+b2 = 8b + 14 = 100
Or, 86×86×86+14×14×14/86×86-86×14+14×14 = 100