Chapter 4: Algebraic Identities Exercise – 4.3
Question: 1
Find the cube of each of the following binomial expression
(a) (1/x + y/3)
(b) (3/x − 2/x2)
(c) (2x + 3/x)
(d) (4 − 1/3x)
Solution:
(a) Given,
(1/x + y/3))3
The above equation is in the form of (a + b)3 = a3 + b3 + 3ab(a + b)
We know that, a = 1/x, b = y/3
By using (a + b)3 formula
(1/x + y/3))3
= (1/x)3 + (y/3)3 + 3(1/x)( y/3)(1/x + y/3)
= 1/x3 + y3/27 + 3 * 1/x * y/3(1/x + y/3)
= 1/x3 + y3/27 + y/x(1/x + y/3)
= 1/x3 + y3/27 + (y/x * 1/x) +(y/x * y3)
= 1/x3 + y3/27 + y/x2 + y2/3x)
Hence,
(1/x + y/3))3 = 1/x3 + y3/27 + y/x2 + y2/3x)
(b) Given,
((3/x−2/x2))3
The above equation is in the form of (a - b)3 = a3 - b3 - 3ab(a - b)
We know that, a = 3/x, b = 2/x2
By using (a - b)3 formula
((3/x − 2/x2))3 = (3x)3 - (2/x2)3 - 3(3/x)( 2/x2)(3/x - 2/x2)
= 27/x3 - 8/x6 - 3 * 3/x * 2/x2(3/x - 2/x2)
= 27/x3 - 8/x6 - 18/x3(3/x - 2/x2)
= 27/x3 - 8/x6 - (18/x3 * 3/x) + (18/x3 * 2/x2)
= 27/x3 - 8/x6 - 54/x4 + 36/x5
Hence, ((3/x−2/x2))3 = 27/x3 - 8/x6 - 54/x4 + 36/x5
(c) Given,
(2x + 3/x)3
The above equation is in the form of (a + b)3 = a3 + b3 + 3ab(a + b)
We know that, a = 2x, b = 3/x
By using (a + b)3 formula
= 8x3 + 27/x3 + 18x/x(2/x + 3/x)
= 8x3 + 27/x3 + 18x/x(2x + 3/x)
= 8x3 + 27/x3 + (18 * 2x) + (18 * 3/x)
= 8x3 + 27/x3 + 36 × 54/x)
Hence,
The cube of (2x + 3/x)3 = 8x3 + 27/x3 + 36 × 54/x)
(d) Given,
(4 − 1/3x)3
The above equation is in the form of (a - b)3 = a3 - b3 - 3ab(a - b)
We know that, a = 4, b = 1/3x
By using (a - b)3 formula
(4 − 1/3x)3 = 43 - (1/3x)3 - 3(4)(1/3x)(4 - 1/3x)
= 64 - 1/27x3 - 12/3x(4 - 1/3x)
= 64 - 1/27x3 - 4/x(4 - 1/3x)
= 64 - 1/27x3 - (4/3x * 4) + (4/3x * 1/3x)
= 64 - 1/27x3 - 16/x + (4/3x2)
Hence,
The cube of (4 − 1/3x)3 = 64 - 1/27x3 - 16/x + (4/3x2)
Question: 2
Simplify each of the following
(a) (x + 3)3 + (x - 3)3
(b) (x/2 + y/3)3 - (x/2 - y/3)3
(c) (x + 2/x)3 + (x − 2/x)3
(d) (2x - 5y)3 - (2x + 5y)3
Solution:
(a) (x + 3)3 + (x – 3)3
The above equation is in the form of a3 + b3 = (a + b)(a2 + b2 – ab)
We know that, a = (x + 3), b = (x – 3)
By using (a3 + b3) formula
= (x + 3 + x – 3)[(x + 3)3 + (x – 3)3 – (x + 3)(x – 3)]
= 2x[(x2 + 32 + 2*x*3) + (x2 + 32 – 2*x*3) – (x2 – 32)]
= 2x[(x2 + 9 + 6x) + (x2 + 9 – 6x) – x2 + 9]
= 2x[(x2 + 9 + 6x + x2 – 9 – 6x – x2 + 9)]
= 2x(x2 + 27)
= 2x3 + 54x
Hence, the result of (x + 3)3 + (x – 3)3 is 2x3 + 54x
(b) (x/2 + y/3)3 - (x/2 – y/3)3
The above equation is in the form of a3 - b3 = (a - b)(a2 + b2 + ab)
We know that, a = (x/2 + y/3)3, b = (x/2 - y/3)3
By using (a3 - b3) formula
= [((x/2 + y/3)3 - ((x/2 − y/3)3)][((x/2 + y/3)3)2((x/2 - y/3)3)2 - ((x/2 + y/3)3)((x/2 - y/3)3)
= (x/3 + y/3 − x/2 + y/3)[((x/2)2 + (y/3)2 + (2xy/6)) + ((x/2)2 + (y/3)2 − (2xy/6)) + ((x/2)2 − (y/3)2)]
= 2y/3[(x2/4 + y2/9 + 2xy/6) + (x2/4 + y2/9 − 2xy/6) + x2/4 − y2/9]
= 2y/3[x2/4 + y2/9 + 2xy/6 + x2/4 + y2/9 − 2xy/6 + x2/4 − y2/9]
= 2y/3[x2/4 + y2/9 + x2/4 + x2/4]
= 2y/3[3x2/4 + y2/9]
= x2y/2 + 2y3/27
Hence, the result of (x/2 + y/3)3 - (x/2 - y/3)3 = x2y/2 + 2y3/27
(c) (x + 2/x)3 + (x − 2/x)3
The above equation is in the form of a3 + b3 = (a + b)(a2 + b2 - ab)
We know that, a = (x + 2x)3, b = (x − 2x)3
By using (a3 + b3) formula
= (x + 2/x + x − 2/x)[(x + 2/x)2 + (x − 2/x)2 − ((x + 2/x)(x − 2/x))]
= (2x)[(x2 + 4/x2 + 4x/x) + (x2 + 4/x2 − 4x/x) − (x2 − 4/x2)
= (2x)[(x2 + 4/x2 + 4x/x + x2 + 4/x2 − 4x/x − x2 + 4/x2)
= (2x)[(x2 + 4/x2 + 4/x2 + 4/x2)
= (2x)[(x2 + 12/x2)
= 2x3 + 24/x
Hence, the result of (x + 2/x)3 + (x − 2/x)3 = (2x)[(x2 + 12/x2)
(d) (2x - 5y)3 - (2x + 5y)3
Given, (2x - 5y)3 - (2x + 5y)3
The above equation is in the form of a3 - b3 = (a - b)(a2 + b2 + ab)
We know that, a = (2x - 5y), b = (2x + 5y)
By using (a3 - b3) formula
= (2x – 5y – 2x – 5y)[(2x – 5y)2 + (2x + 5y)2 + ((2x – 5y) * (2x + 5y))]
= (-10y)[(4x2 + 25y2 – 20xy) + (4x2 + 25y2 + 20xy) + 4x2 – 25y2]
= (-10y)[ 4x2 + 25y2 – 20xy + 4x2 + 25y2 + 20xy + 4x2 – 25y2]
= (-10y)[4x2 + 4x2 + 4x2 + 25y2]
= (-10y)[12x2 + 25y2}
= -120x2y – 250y3
Hence, the result of (2x – 5y)3 – (2x + 5y)3 = -120x2y – 250y3
Question: 3
If a + b = 10 and ab = 21, Find the value of a3 + b3
Solution:
Given,
a + b = 10, ab = 21
we know that, (a + b)3 = a3 + b3 + 3ab(a + b) ... 1
substitute a + b = 10 , ab = 21 in eq 1
⟹ (10)3 = a3 + b3 + 3(21)(10)
⟹ 1000 = a3 + b3 + 630
⟹ 1000 – 630 = a3 + b3
⟹ 370 = a3 + b3
Hence, the value of a3 + b3 = 370
Question: 4
If a - b = 4 and ab = 21, Find the value of a3 - b3
Solution:
Given,
a - b = 4, ab = 21
we know that, (a - b)3 = a3 - b3 - 3ab(a - b) -------- 1
substitute a - b = 4 , ab = 21 in eq 1
⟹ (4)3 = a3 - b3 - 3(21)(4)
⟹ 64 = a3 - b3 - 252
⟹ 64 + 252 = a3 - b3
⟹ 316 = a3 - b3
Hence, the value of a3 - b3 = 316
Question: 5
If (x + 1/x) = 5, Find the value of x3 + 1/x3
Solution:
Given, (x + 1/x) = 5
We know that, (a + b)3 = a3 + b3 + 3ab(a + b) ... 1
Substitute (x + 1/x) = 5 in eq1
(x + 1/x)3 = x3 + 1/x3 + 3(x * 1/x)(x + 1/x)
53 = x3 + 1/x3 + 3(x * 1/x)(x + 1/x)
125 = x3 + 1/x3 + 3(x + 1/x)
125 = x3 + 1/x3 + 3(5)
125 = x3 + 1/x3 + 15
125 – 15 = x3 + 1/x3
x3 +1/x3 = 110
Hence, the result is x3 + 1/x3 = 110
Question: 6
If (x − 1/x) = 7, Find the value of x3 − 1/x3
Solution:
Given, If (x − 1/x) = 7
We know that, (a - b)3 = a3 - b3 - 3ab(a - b) ... 1
Substitute (x − 1/x) = 7 in eq 1
(x − 1/x)3 = x3 − 1/x3 - 3(x * 1/x)(x - 1/x)
73 = x3 - 1/x3 - 3(x - 1/x)
343 = x3 -1/x3 - (3 * 7)
343 = x3 - 1/x3 - 21
343 + 21 = x3 - 1/x3
x3 - 1/x3 = 364
hence, the result is x3 - 1/x3 = 364
Question: 7
If (x − 1/x) = 5, Find the value of x3 − 1/x3
Solution:
Given, If (x − 1/x) = 5
We know that, (a - b)3 = a3 - b3 - 3ab(a - b) ... 1
Substitute (x − 1/x) = 5 in eq 1
(x − 1/x)3 = x3 - 1/x3 - 3(x * 1/x)(x - 1/x)
53 = x3 - 1/x3 - 3(x - 1/x)
125 = x3 - 1/x3 - (3 * 5)
125 = x3 - 1/x3 - 15
125 + 15 = x3 - 1/x3
x3 - 1/x3 = 140
Hence, the result is x3 - 1/x3 = 140
Question: 8
If (x2 + 1/x2) = 51, Find the value of x3 − 1/x3
Solution:
Given, (x2 + 1/x2) = 51
We know that, (x - y)2 = x2 + y2 - 2xy .... 1
Substitute (x2 + 1/x2) = 51 in eq 1
(x - 1/x)2 = x2 + 1/x2 - 2 * x * 1/x
(x - 1/x)2 = x2 + 1/x2 - 2
(x - 1/x)2 = 51 - 2
(x - 1/x)2 = 49
(x – 1/x) =√49
(x - 1/x) = ±7
We need to find x3 − 1/x3
So, a3 - b3 = (a - b)(a2 + b2 + ab)
x3 − 1/x3 = (x - 1/x)(x2 + 1/x2 + (x * 1/x)
We know that,
(x -1/x) = 7 and (x2 + 1/x2) = 51
x3 − 1/x3 = 7(51 + 1)
x3 − 1/x3 = 7(52)
x3 − 1/x3 = 364
Hence, the value of x3 − 1/x3 = 364
Question: 9
If (x2 + 1/x2) = 98, Find the value of x3 + 1/x3
Solution:
Given, (x2 + 1/x2) = 98
We know that, (x + y)2 = x2 + y2 + 2xy ... 1
Substitute (x2 + 1/x2) = 98 in eq 1
(x + 1/x)2 = x2 + 1/x2 + 2 * x * 1/x
(x + 1/x)2 = x2 + 1/x2 + 2
(x + 1/x)2 = 98 + 2
(x + 1/x)2 = 100
(x + 1/x) = √100
(x + 1/x) = ± 10
We need to find x3 + 1/x3
So, a3 + b3 = (a + b)(a2 + b2 – ab)
x3 + 1/x3 = (x + 1/x)(x2 + 1/x2 - (x * 1/x)
We know that,
(x + 1/x) = 10 and (x2 + 1/x2) = 98
x3 + 1/x3 = 10(98 - 1)
x3 + 1/x3 = 10(97)
x3 + 1/x3 = 970
Hence, the value of x3 + 1/x3 = 970
Question: 10
If 2x + 3y = 13 and xy = 6, Find the value of 8x3 + 27y3
Solution:
Given, 2x + 3y = 13, xy = 6
We know that,
(2x + 3y)3 = 132
⟹ 8x3 + 27y3 + 3(2x)(3y)(2x + 3y) = 2197
⟹ 8x3 + 27y3 + 18xy(2x + 3y) = 2197
Substitute 2x + 3y = 13, xy = 6
⟹ 8x3 + 27y3 + 18(6)(13) = 2197
⟹ 8x3 + 27y3 + 1404 = 2197
⟹ 8x3 + 27y3 = 2197 – 1404
⟹ 8x3 + 27y3 = 793
Hence, the value of 8x3 + 27y3 = 793
Question: 11
If 3x - 2y = 11 and xy = 12, Find the value of 27x3 - 8y3
Solution:
Given, 3x - 2y = 11, xy = 12
We know that (a – b)3 = a3 – b3 – 3ab(a + b)
(3x - 2y)3 = 113
⟹ 27x3 – 8y3 – (18 * 12 * 11) = 1331
⟹ 27x3 – 8y3 – 2376 = 1331
⟹ 27x3 – 8y3 = 1331 + 2376
⟹ 27x3 – 8y3 = 3707
Hence, the value of 27x3 – 8y3 = 3707
Question: 12
If x4 + (1/x4) = 119, Find the value of x3 − (1/x3)
Solution:
Given, x4 + (1/x4) = 119 .... 1
We know that (x + y)2 = x2 + y2 + 2xy
Substitute x4 + (1/x4) = 119 in eq 1
(x2 + (1/x2))2 = x4 + (1/x4) + (2*x2* 1/x2)
= x4 + (1/x4) + 2
= 119 + 2
= 121
(x2 + (1/x2))2 = 121
x2 + (1/x2) = ±11
Now, find (x - 1/x)
We know that (x - y)2 = x2 + y2 - 2xy
(x - 1/x)2 = x2 + 1/x2 - (2*x*1/x
= x2 + 1/x2 - 2
= 11- 2
= 9
(x – 1/x) = √9
= ±3
We need to find x3 − (1/x3)
We know that, a3 - b3 = (a - b)(a2 + b2 - ab)
x3 − (1/x3) = (x - 1/x)(x2 + (1/x2) + x * 1/x
Here, x2 + (1/x2) = 11 and (x - 1/x) = 3
x3 − (1/x3) = 3(11 + 1)
= 3(12)
= 36
Hence, the value of x3 − (1/x3) = 36
Question: 13
Evaluate each of the following
(a) (103)3
(b) (98)3
(c) (9.9)3
(d) (10.4)3
(e) (598)3
(f) (99)3
Solution:
Given,
(a) (103)3
we know that (a + b)3 = a3 + b3 + 3ab(a + b)
⟹ (103)3 can be written as (100 + 3)3
Here, a = 100 and b = 3
(103)3 = (100 + 3)3
= (100)3 + (3)3 + 3(100)(3)(100 + 3)
= 1000000 + 27 + (900*103)
= 1000000 + 27 + 92700
= 1092727
The value of (103)3 = 1092727
(b) (98)3
we know that (a - b)3 = a3 - b3 - 3ab(a - b)
⟹ (98)3 can be written as (100 - 2)3
Here, a = 100 and b = 2
(98)3 = (100 - 2)3
= (100)3 - (2)3 - 3(100)(2)(100 - 2)
= 1000000 - 8 - (600*102)
= 1000000 – 8 – 58800
= 941192
The value of (98)3 = 941192
(c) (9.9)3
we know that (a - b)3 = a3 - b3 - 3ab(a - b)
⟹ (9.9)3 can be written as (10 – 0.1)3
Here, a = 10 and b = 0.1
(9.9)3 = (10 – 0.1)3
= (10)3 - (0.1)3 - 3(10)(0.1)(10 – 0.1)
= 1000 – 0.001 - (3*9.9)
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
The value of (9.9)3 = 970.299
(d) (10.4)3
we know that (a + b)3 = a3 + b3 + 3ab(a + b)
⟹ (10.4)3 can be written as (10 + 0.4)3
Here, a = 10 and b = 0.4
(10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3(10)(0.4)(10 + 0.4)
= 1000 + 0.064 + (12*10.4)
= 1000 + 0.064 + 124.8
= 1000 + 124.864
= 1124.864
The value of (10.4)3 = 1124.864
(e) (598)3
we know that (a - b)3 = a3 - b3 - 3ab(a - b)
⟹ (598)3 can be written as (600 - 2)3
Here, a = 600 and b = 2
(598)3 = (600 - 2)3
= (600)3 - (2)3 - 3(600)(2)(600 - 2)
= 216000000 - 8 - (3600*598)
= 216000000 - 8 - 2152800
= 216000000 - 2152808
= 213847192
The value of (598)3 = 213847192
(f) (99)3
we know that (a - b)3 = a3 - b3 - 3ab(a - b)
⟹ (99)3 can be written as (100 - 1)3
Here , a = 100 and b = 1
(99)3 = (100 - 1)3
= (100)3 - (1)3 - 3(100)(1)(100 - 1)
= 1000000 - 1 - (300*99)
= 1000000 - 1 - 29700
= 1000000 - 29701
= 970299
The value of (99)3 = 970299
Question: 14
Evaluate each of the following
(a) 1113 - 893
(b) 463 + 343
(c) 1043 + 963
(d) 933 - 1073
Solution:
(a) Given,
1113 - 893
the above equation can be written as (100 + 11)3 - (100 - 11)3
we know that, (a + b)3 - (a - b)3 = 2[b3 + 3ab2]
here, a = 100 b = 11
(100 + 11)3 - (100 - 11)3 = 2[113 + 3(100)2(11)]
= 2[1331 + 330000]
= 2[331331]
= 662662
The value of 1113 - 893 = 662662
(b) 463 + 343
the above equation can be written as (40 + 6)3 + (40 - 6)3
we know that, (a + b)3 + (a - b)3 = 2[a3 + 3ab2]
here, a= 40 , b = 4
(40 + 6)3 + (40 - 6)3 = 2[403 + 3(6)2(40)]
= 2[64000 + 4320]
= 2[68320]
= 1366340
The value of 463 + 343 = 1366340
(c) 1043 + 963
the above equation can be written as (100 + 4)3 + (100 - 4)3
we know that, (a + b)3 + (a - b)3 = 2[a3 + 3ab2]
here, a= 100 b = 4
(100 + 4)3 - (100 - 4)3 = 2[1003 + 3(4)2(100)]
= 2[1000000 + 4800]
= 2[1004800]
= 2009600
The value of 1043 + 963 = 2009600
(d) 933 - 1073
the above equation can be written as (100 - 7)3 - (100 + 7)3
we know that, (a - b)3 - (a + b)3 = -2[b3 + 3ba2]
here, a = 93, b = 107
(100 - 7)3 - (100 + 7)3 = - 2[73 + 3(100)2(7)]
= - 2[343 + 210000]
= - 2[210343]
= - 420686
The value of 933 - 1073 = - 420686
Question: 15
If x + 1/x = 3, calculate x2 + 1/x2, x3 + 1/x3, x4 + 1/x4
Solution:
Given, x + 1/x = 3
We know that (x + y)2 = x2 + y2 + 2xy
(x + 1/x)2 = x2 + 1/x2 + (2∗x∗1/x)
32 = x2 + 1/x2 + 2
9 - 2 = x2 + 1/x2
x2 + 1/x2 = 7
Squaring on both sides
(x2 + 1/x2)2 = 72
x4 + 1/x4 + 2* x2 * 1/x2 = 49
x4 + 1/x4 + 2 = 49
x4 + 1/x4 = 49 - 2
x4 + 1/x4 = 47
Again, cubing on both sides
(x + 1/x)3 = 33
x3 + 1/x3 + 3x*1/x(x + 1/x) = 27
x3 + 1/x3 + (3*3) = 27
x3 + 1/x3 + 9 = 27
x3 + 1/x3 = 27 - 9
x3 + 1/x3 = 18
Hence, the values are x2 + 1/x2 = 7, x4 + 1/x4 = 47, x3 + 1/x3 = 18
Question: 16
If x4 + 1/x4 = 194, calculate x2 + 1/x2, x3 + 1/x3, x + 1/x
Solution:
Given,
x4 + 1/x4 = 194 ... 1
add and subtract (2*x2∗1/x2) on left side in above given equation
x4 + 1/x4 + (2*x2∗1/x2) - 2(2*x2∗1/x2) = 194
x4 + 1/x4 + (2*x2∗1/x2) - 2 = 194
(x2 + 1/x2)2 - 2 = 194
(x2 + 1/x2)2 = 194 + 2
(x2 + 1/x2)2 = 196
(x2 + 1/x2) = 14 ... 2
Add and subtract (2*x* 1/x) on left side in eq 2
(x2 + 1/x2) + (2*x* 1/x) - (2*x* 1/x) = 14
(x + 1/x)2 - 2 = 14
(x + 1/x)2 = 14 + 2
(x + 1/x)2 = 16
(x + 1/x) = √16
(x + 1/x) = 4 ... 3
Now, cubing eq 3 on both sides
(x + 1/x)3 = 43
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
x3 + 1/x3 + 3*x*1/x(x + 1/x) = 64
x3 + 1/x3 + (3*4) = 64
x3 + 1/x3 = 64 - 12
x3 + 1/x3 = 52
hence, the values of (x2 + 1/x2)2 = 196, (x + 1/x) = 4, x3 + 1/x3 = 52
Question: 17
Find the values of 27x3 + 8y3, if
(a) 3x + 2y = 14 and xy = 8
(b) 3x + 2y = 20 and xy = 14/9
Solution:
(a) Given, 3x + 2y = 14 and xy = 8
cubing on both sides
(3x + 2y)3 = 143
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
27x3 + 8y3 + 3(3x)(2y)(3x + 2y) = 2744
27x3 + 8y3 + 18xy(3x + 2y) = 2744
27x3 + 8y3 + 18(8)(14) = 2744
27x3 + 8y3 + 2016 = 2744
27x3 + 8y3 = 2744 - 2016
27x3 + 8y3 = 728
Hence, the value of 27x3 + 8y3 = 728
(b) Given, 3x + 2y = 20 and xy = 14/9
cubing on both sides
(3x + 2y)3 = 203
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
27x3 + 8y3 + 3(3x)(2y)(3x + 2y) = 8000
27x3 + 8y3 + 18xy(3x + 2y) = 8000
27x3 + 8y3 + 18(14/9)(20) = 8000
27x3 + 8y3 + 560 = 8000
27x3 + 8y3 = 8000 - 560
27x3 + 8y3 = 7440
Hence, the value of 27x3 + 8y3 = 7440
Question: 18
Find the value of 64x3 - 125z3, if 4x - 5z = 16 and xz = 12
Solution:
Given, 64x3 - 125z3
Here, 4x - 5z = 16 and xz = 12
Cubing 4x - 5z = 16 on both sides
(4x - 5z)3 = 163
We know that, (a - b)3 = a3 - b3 - 3ab(a - b)
(4x)3 - (5z)3 - 3(4x)(5z)(4x - 5z) = 163
64x3 - 125z3 - 60(xz)(16) = 4096
64x3 - 125z3 - 60(12)(16) = 4096
64x3 - 125z3 - 11520 = 4096
64x3 - 125z3 = 4096 + 11520
64x3 - 125z3 = 15616
The value of 64x3 - 125z3 = 15616
Question: 19
Solution:
Given,
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
