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)