Chapter 4: Algebraic Identities Exercise – 4.4

Question: 1

Find the following products

(a) (3x + 2y)(9x² - 6xy + 4y²)

(b) (4x - 5y)(16x² + 20xy + 25y²)

(c) (7p⁴ + q)(49p⁸ - 7p⁴q + q²)

(d) (x/2 + 2y)(x²/4 - xy + 4y²)

(e) (3/x − 5/y)(9/x² + 25/y² + 15/xy)

(f) (3 + 5/x)(9 − 15/x + 25/x²)

(g) (2/x + 3x)(4/x² + 9x² − 6)

(h) (3/x − 2x²)(9/x² + 4x⁴ − 6x)

(i) (1 - x)(1 + x + x²)

(j) (1 + x)(1 - x + x²)

(k) (x² − 1)(x⁴ + x² + 1)

(l) (x² + 1)(x⁶ − x³ + 1)

Solution:

(a) Given,

(3x + 2y)(9x² - 6xy + 4y²)

We know that,

a³ + b³ = (a + b)(a² + b² - ab)(3x + 2y)(9x² – 6xy + 4y²)

can we written as

⟹ (3x + 2y)[(3x)² – (3x)(2y) + (2y)²)]

⟹ (3x)³ + (2y)³ 

⟹ 27x³ + 8y³ 

Hence, the value of (3x + 2y)(9x² – 6xy + 4y²)

= 27x³ + 8y³(b)(4x – 5y)(16x² + 20xy + 25y²)

(b) Given,

(4x - 5y)(16x² + 20xy + 25y²)

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(4x - 5y)(16x² + 20xy + 25y²)

can we written as

⟹ (4x - 5y)[(4x)² + (4x)(5y) + (5y)²)]

⟹ (4x)³ - (5y)³ 

⟹ 16x³ – 25y³ 

Hence, the value of (4x – 5y)(16x² + 20xy + 25y²)

= 16x³ – 25y³ 

(c) Given,

(7p⁴ + q)(49p⁸ – 7p⁴q + q²)

We know that,

a³ + b³ = (a + b)(a² + b² – ab)(7p⁴ + q)(49p⁸ – 7p⁴q + q²)

can be written as

⟹ (7p⁴ + q)[(7p⁴)² - (7p⁴)(q) + (q)²)]

⟹ (7p⁴)³ + (q)³ 

⟹ 343p¹² + q³ 

Hence, the value of (7p⁴ + q)(49p⁸ – 7p⁴q + q²)

= 343p¹² + q³ 

(d) Given,

(x/2 + 2y)(x²/4 – xy + 4y²)

We know that,

a³ + b³= (a + b)(a² + b² – ab)(x/2 + 2y)(x²/4 – xy + 4y²)

can be written as

⟹ (x/2 + 2y)[(x/2)² − x/2(2y) + (2y)²] 

⟹ (x/2)³ + (2y)³ 

⟹ x³/8 + 8y³ 

(e) Given, 

(3/x − 5/y)(9/x² + 25/y² + 15/xy) 

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(3/x − 5/y)(9/x² + 25/y² + 15/xy) 

Can be written as,

⟹ (3/x − 5/y)(3/x)² + (5/y)² + (3/x)(5/y) 

⟹ (3/x)³ − (5/y)³ 

⟹ (27/x³) − (125/y³) 

Hence, the value of (3/x − 5/y)(9/x² + 25/y² + 15/xy) 

= (27/x³) − (125/y³) 

(f)  Given, 

(3 + 5/x)(9 − 15/x + 25/x²) 

We know that,

a³ + b³ = (a + b)(a² + b² - ab)(3 + 5/x)(9 − 15/x + 25/x²) 

can be written as,

⟹ (3 + 5/x)[(3²) − 3(5/x) + (5/x)²]

⟹ (3)³ + (5/x)³ 

⟹ 27 + 125/x³ 

Hence, the value of (3 + 5x)(9 − 15/x + 25/x²) is 27 + 125/x³ 

(g) Given, 

(2/x + 3x)(4/x² + 9/x² − 6)

We know that,

a³ + b³ = (a + b)(a² + b² - ab)(2/x + 3x)(4/x² + 9/x² − 6) 

can be written as,

⟹ (2/x + 3x)[(2/x)² + (3x)² − (2/x)(3x)] 

⟹ (2/x)³ + (3x)³ 

⟹ 8/x³ + 9x³ 

Hence, the value of (2/x + 3x)(4/x² + 9x² − 6) is 8/x³ + 9x³ 

(h) (3/x − 2x²)(9/x² + 4x⁴ − 6x) 

Given, 

(3/x − 2x²)(9/x² + 4x⁴ − 6x) 

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(3/x − 2x²)(9/x² + 4x⁴ − 6x) 

can be written as,

⟹ (3/x − 2x²)[(3/x)² + (2x²)² − (3/x)(2x²)] 

⟹ (3/x − 2x²)[(9/x²) + 4x⁴ − (3/x)(2x²)] 

⟹ (3/x)³ − (2x²)³ 

⟹ 27/x³ − 8x⁶ 

Hence, (3/x − 2x²)(9/x² + 4x⁴ − 6x) is 27/x³ − 8x⁶ 

(i) (1 - x)(1 + x + x²)

Sol:

Given, (1 - x)(1 + x + x²)

We know that, a³ - b³ = (a - b)(a² + b² + ab)(1 - x)(1 + x + x²)

can be written as, ⟹ (1 - x)[(1² + (1)(x)+ x²)]

⟹ (1)³ - (x)³ 

⟹ 1 – x³

Hence, the value of (1 – x)(1 + x + x²) is 1 – x³ 

(j) Given,

(1 + x)(1 – x + x²)

We know that,

a³ + b³ = (a + b)(a² + b² – ab)(1 + x)(1 – x + x²)

can be written as,

⟹ (1 + x)[(1² - (1)(x) + x²)]

⟹ (1)³ + (x)³ 

⟹ 1 + x³ 

Hence, the value of (1 + x)(1 + x - x²) is 1 + x³ (k)(x² − 1)(x⁴ + x² + 1)

(k) Given, 

(x² − 1)(x⁴ + x² + 1) 

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(x² − 1)(x⁴ + x² + 1) 

can be written as,

⟹ (x² - 1)[(x²)² - 1² + (x²)(1)]

⟹ (x²)³ - 1³ 

⟹ x⁶ - 1

Hence, (x² − 1)(x⁴ + x² + 1) is x⁶ - 1

(l) Given, 

(x² + 1)(x⁶ − x³ + 1) 

We know that, a³ + b³ = (a + b)(a² + b² - ab)(x² + 1)(x⁶ − x³ + 1) 

can be written as, ⟹ (x³ + 1)[(x³)² - (x³)(1) + 1²]

⟹ (x³)³ + 1³ 

⟹ x⁹ + 1

Hence, the value of (x² + 1)(x⁶ − x³ + 1) is x⁹ + 1

Question: 2

Find x = 3 and y = -1, Find the values of each of the following using in identity:

(a) (9x² - 4x²)(81y⁴ + 36x²y² + 16x⁴)

(b) (3/x − 5/y)(9/x² + 25/y² + 15/xy)

(c) (x/7 + y/3)(x²/49 + y²/9 − xy/21)

(d) (x/4 − y/3)(x²/16 + y²/9 + xy/21)

(e) (5/x + 5x)(25/x² − 25 + 25x²)

Solution:

(a)  We know that,

a³ - b³ = (a - b)(a² + b² + ab)

(9x² – 4x²)(81y⁴ + 36x²y² + 16x⁴) can be written as,

⟹ (9x² – 4x²)[(9y²)² + (9)(4)x²y² + (4x²)²]

⟹ (9y²)³ – (4x²)³ 

⟹ 729y⁶ – 64x⁶ 

Substitute the value x = 3, y = -1 in 729y⁶ – 64x⁶ we get,

⟹ 729y⁶ – 64x⁶ 

⟹ 729(-1)⁶ – 64(3)⁶ 

⟹ 729(1) – 64(729)

⟹ 729 – 46656

⟹ -45927

Hence, the product value of (9x² – 4x²)(81y⁴ + 36x²y² + 16x⁴) = -45927

(b) Given, (3/x − 5/y)(9/x² + 25/y² + 15/xy)

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(3/x − 5/y)(9/x² + 25/y² + 15/xy) 

Can be written as, ⟹ (3/x − x/3)[(3/x)² + (x/3)² + (3/x)(x/3)] 

⟹ (3/x)³ − (x/3)³ 

⟹ (27/x³) − (x³/27) ... 1

Substitute x = 3 in eq 1

⟹ (27/3³) − (3³/27) 

⟹ (27/27) − (27/27) 

⟹ 0

Hence, the value of (3/x − 5/y)(9/x² + 25/y² + 15/xy) is 0

(c) Given,

(x/7 + y/3)(x²/49 + y²/9 − xy/21) 

 We know that,

a³ + b³ = (a + b)(a² + b² - ab)(x/7 + y/3)(x²/49 + y/29 − xy/21) 

Can be written as,

⟹ (x/7 + y/3)[(x/7)² + (y/3)² − (x/7)(y/3)] 

⟹ (x/7)³ + (y/3)³ 

⟹ (x³/343) + (y³/27) ... 1

Substitute x = 3, y = -1 in eq 1

⟹ (3³/343) + ((−1)³/27) 

⟹ (27/343) − (1/27) 

Taking least common multiple, we get

Hence, the value of (x/7 + y/3)(x²/49 + y²/9 − xy/21)

= 386/9261 

(d) Given, 

(x/4 − y/3)(x²/16 + y²/9 − +xy/21) 

We know that,

a³ - b³ = (a - b)(a² + b² + ab)

(x⁴ − y³)(x²/16 + y²/9 + xy/21) 

Can be written as,

⟹ (x/4 − y/3)[(x/4)² + (y/3)² + (x/4)(y/3)] 

⟹ (x/4)³ − (y/3)³ 

⟹ (x³/64) − (y³/27) .... 1

Substitute x = 3, y = -1 in eq 1

⟹ (3³/343) − ((−1)³/27) 

⟹ (27/64) + (1/27) 

Taking least common multiple, we get

⟹ 793/9261

Hence, the value of (x/4 − y/3)(x²/16 + y²/9 + xy/21) 

= 793/1728

(e) Given,

(5/x + 5x)(25/x² − 25 + 25x²) 

We know that,

a³ + b³ = (a + b)(a² + b² - ab)

(5/x + 5x)(25/x² − 25 + 25x²) 

can be written as, ⟹ (5/x + 5x)[(5/x)² + (5x)² − (5/x)(5x)] 

⟹ (5/x)³ + (5x)³ 

⟹ 125/x³ + 125x³ ... 1

Substitute x = 3, in eq 1

⟹ 125/3³ + 125(3)³ 

⟹ 125/27 + 125∗27 

⟹ 125/27 + 3375

Taking least common multiple, we get

Hence, the value of (5/x + 5x)(25/x² − 25 + 25x²) is 91250/25

Question: 3

If a + b = 10 and ab = 16, find the value of a² - ab + b² and a² + ab + b²

Solution:

Given,

a + b = 10, ab = 16

We know that,

(a + b)³ = a³ + b³ + 3ab(a + b)

⟹ a³ + b³ = (a + b)³ - 3ab(a + b)

⟹ a³ + b³ = (10)³ – 3(16)(10)

⟹ a³ + b³ = 1000 – 480

⟹ a³ + b³ = 520

Substitute, a³ + b³ = 520,

a + b = 10 in a³ + b³ 

= (a + b)(a²+ b² – ab) a³ + b³ 

= (a + b)(a² + b² – ab) 520

= 10(a² + b² – ab) 520/10 

= (a² + b² – ab)

⟹ (a² + b² – ab) = 52

Now, we need to find (a² + b² + ab)

Add and subtract 2ab in a² + b² + ab

⟹ a² + b² + ab – 2ab + 2ab

⟹ (a + b)² - ab

Substitute a + b = 10, ab

⟹ a² + b² + ab = 10² – 16 = 100 – 16 = 84

Hence, the values of (a² + b²– ab) = 52 and (a² + b² + ab) = 84

Question: 4

If a + b = 8 and ab = 6, find the value of a³ + b³

Solution:

Given,

a + b = 8 and ab = 6

We know that,

a³ + b³ = (a + b)³ – 3ab(a + b)

⟹ a³ + b³ = (a + b)³ – 3ab(a + b)

⟹ a³ + b³ = (8)³ – 3(6)(8)

⟹ a³ + b³ = 512 – 144

⟹ a³ + b³ = 368

Hence, the value of a³ + b³ is 368

Question: 5

If a – b = 6 and ab = 20, find the value of a³ - b³

Solution:

Given,

a – b = 6 and ab = 20 

We know that,

a³ - b³ = (a - b)³ + 3ab(a - b)

⟹ a³ - b³ = (a - b)³ + 3ab(a - b)

⟹ a³ - b³ = (6)³ + 3(20)(6)

⟹ a³ - b³ = 216 + 360

⟹ a³ - b³ = 576

Hence, the value of a³ - b³ is 576

Question: 6

If x = – 2 and y = 1, by using an identity find the value of the following:

(a) (4y² - 9x²)(16y⁴ + 36x²y² + 81x⁴)

(b) (2/x − x/2)(4/x² + x²/4 + 1)

(c) (5y + 15/y)(25y² − 75 + 225/y²)

Solution:

Given,

(a) (4y² - 9x²)(16y⁴ + 36x²y² + 81x⁴)

We know that,

a³ - b³ = (a - b)(a² + b² + ab)(4y² – 9x²)(16y⁴ + 36x²y² + 81x⁴)

can be written as,

⟹ (4y² – 9x²)[(4x)² + 4y²*9x² + (9x²)²)

⟹ (4y²)³ – (9x²)³ 

⟹ 64y⁶ – 729x⁶ ... 1

Substitute x = -2 and y = 1 in eq 1

⟹ 64y⁶ – 729x⁶ 

⟹ 64(1)⁶ – 729(-2)⁶ 

⟹ 64 – 729(64)

⟹ 64(1 – 729)

⟹ 64(-728)

⟹ – 46592

Hence, the value of (4y² – 9x²)(16y⁴ + 36x²y² + 81x⁴) is – 46592

(b) (2/x − x/2)(4/x² + x²/4 + 1) here x = -2

We know that,

a³ - b³= (a - b)(a² + b² + ab) (2/x − x/2)(4/x² + x²/4 + 1) 

can be witten as,

⟹ (2/x − x/2)[(2/x)² + (x/2)² + (2/x)(x/2)] 

⟹ (2/x)³ − (x/2)³ 

⟹ (8/x³) − (x³/8) ... 1

Substitute x = -2 in eq 1

⟹ (8/(−2)³) − ((−2)³/8) 

⟹ (8/−8) − (−8/8) 

⟹ -1 + 1

⟹ 0

Hence, the value of (2/x − x/2)(4/x² + x²/4 + 1) is 0

(c) (5y + 15/y)(25y² − 75 + 225/y²) 

We know that,

a³ + b³ = (a + b)(a² + b² - ab)

(5y + 15/y)(25y² − 75 + 225/y²) 

can be written as,

⟹ (5y + 15/y)[(5y)² + (15y)² - (5y)( 15/y)]

⟹ (5y)³ + (15/y)³ 

⟹ 125y³ + (3375/y³) ... 1

Substitute y = 1 in eq 1

⟹ 125(1)³ + (3375/(1)²)

⟹ 125 + 3375

⟹ 3500

Hence, the value of (5y + 15/y)(25y² − 75 + 225/y²) is 3500.

Question: 7

Find the following products:

(a) (3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx)

(b) (4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx)

(c) (2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca)

(d) (3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz)

Solution:

Given,

(a) (3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx)

we know that,

x³+ y³ + z³ - 3xyz

= (x + y + z)(x² + y² + z² - xy - yz - zx) so, (3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx)

= (3x)³ + (2y)³ + (2z)³ - 3(3x)(2y)(2z)

= 27x³ + 8y³ + 8z³ - 36xyz

Hence, the value of (3x + 2y + 2z)(9x² + 4y² + 4z² - 6xy - 4yz - 6zx) is 27x³ + 8y³ + 8z³ - 36xyz

(b) (4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx)

we know that,

x³ + y³ + z³ - 3xyz

= (x + y + z)(x² + y² + z² - xy - yz - zx) so, (4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx)

= (4x)³ + (-3y)³ + (2z)³ -3(4x)(-3y)(2z)

= 64x³ - 27y³+ 8z³ + 72xyz

Hence, the value of (4x - 3y + 2z)(16x² + 9y² + 4z² + 12xy + 6yz - 8zx) is 64x³ - 27y³+ 8z³ + 72xyz

(c) (2a - 3b - 2c)(4a²+ 9b² + 4c² + 6ab - 6bc + 4ca)

we know that,

x³ + y³ + z³ - 3xyz

= (x + y + z)(x²+ y² + z² - xy - yz - zx) so, (2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca)

= (2a)³ + (-3b)³ + (-2c)³ - 3(2a)(-3b)(-2c)

= 8a³ - 27b³ - 8c³ - 36abc

Hence, the value of

(c) (2a - 3b - 2c)(4a² + 9b² + 4c² + 6ab - 6bc + 4ca) is 8a³ - 27b³ - 8c³ - 36abc

(d) (3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz)

we know that, x³ + y³ + z³ - 3xyz

= (x + y + z)(x² + y² + z² - xy - yz - zx) so, (3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz)

= (3x)³ + (-4y)³ + (5z)³ -3(3x)(-4y)(5z) = 27x³ - 64y³ + 125z³ + 180xyz

Hence, the value of (3x - 4y + 5z)(9x² + 16y² + 25z² + 12xy - 15zx + 20yz) is 27x³ - 64y³ + 125z³ + 180xyz

Question: 8

If x + y + z = 8 and xy + yz + zx = 20, Find the value of x³ + y³ + z³ - 3xyz

Solution:

Given,

x + y + z = 8 and xy + yz + zx = 20

We know that,

(x + y + z)² = x² + y²+ z² + 2(xy + yz + zx)

(x + y + z)² = x² + y² + z² + 2(20)

(x + y + z)² = x² + y² + z²+ 40

8² = x² + y² + z² + 40

64 - 40 = x² + y² + z² 

x² + y² + z² = 24

we know that, x³ + y³ + z³ - 3xyz = (x + y + z)

(x² + y² + z² - xy - yz - zx) x³ + y³ + z³ - 3xyz

= (x + y + z)[(x² + y² + z²) - (xy + yz + zx)] here,

x + y + z = 8,

xy + yz + zx = 20,

x²+ y² + z² = 24

x³ + y³ + z³ - 3xyz = 8[(24 - 20)] = 8 * 4 = 32

Hence, the value of x³ + y³ + z³ - 3xyz is 32

Question: 9

If a + b + c = 9 and ab + bc + ca = 26, Find the value of a³ + b³ + c³ - 3abc

Solution:

Given,

a + b + c = 9 and ab + bc + ca = 26

We know that,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b + c)² = a² + b² + c² + 2(26)

(a + b + c)² = a² + b²+ c² + 52

9² = a² + b² + c² + 52

81 - 52 = a² + b² + c² 

a² + b² + c² = 29

we know that,

a³ + b³ + c³ - 3abc = (a + b + c)

(a² + b² + c² - ab - bc - ca) a³ + b³ + c³ - 3abc = (a + b + c)[(a² + b² + c²) - (ab + bc + ca)] here,

a + b + c = 9,

ab + bc + ca = 26,

a² + b² + c² = 29

a³ + b³ + c³ - 3abc = 9[(29 - 26)] = 9 * 3 = 27

Hence, the value of a³ + b³ + c³ - 3abc is 27

Question: 10

If a + b + c = 9 and a² + b² + c² = 35, Find the value of a³ + b³ + c³ - 3abc

Solution:

Given,

a + b + c = 9 and a² + b² + c² = 35

We know that,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

9² = 35 + 2(ab + bc + ca)

81 = 35 + 2(ab + bc + ca)

81 - 35 = 2(ab + bc + ca) 

46/2 = ab + bc + ca

ab + bc + ca = 23

we know that, a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

a³ + b³ + c³ - 3abc = (a + b + c)[(a² + b² + c²) - (ab + bc + ca)] here,

a + b + c = 9,

ab + bc + ca = 23,

a² + b² + c² = 35 a³ + b³ + c³ - 3abc = 9[(35 - 23)] = 9 * 12 = 108

Hence, the value of a³ + b³ + c³ - 3abc is 108

Question: 11

Evaluate:

(a) 25³ - 75³ + 50³

(b) 48³ - 30³ - 18³

(c) (1/2)³ + (1/3)³ − (5/6)³

(d) (0.2)³ - (0.3)³ + (0.1)³

Solution:

Given,

(a) 25³ - 75³ + 50³ 

we know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a²+ b² + c² - ab - bc - ca) here,

a = 25, b = -75, c = 50

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

a³ + b³ + c³ = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (25 - 75 + 50)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (0)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = 3abc

25³ + (-75)³+ 50³ = 3abc

= 3(25)(-75)(50)

= – 281250

Hence, the value 25³ + (- 75)³ + 50³ = – 281250

(b) 48³ - 30³ - 18³ 

we know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

here, a = 48, b = -30, c = -18

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

a³ + b³ + c³ = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (48 - 30 - 18)

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (0)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = 3abc

48³ + (-30)³+ (-18)³ = 3abc

= 3(48)(-30)(-18) = 77760

Hence, the value 48³ + (-30)³ + (-18)³ = 77760

(c) (1/2)³ + (1/3)³ − (5/6)³ 

we know that,

a³ + b³ + c³ - 3abc

= (a + b + c)(a² + b² + c² - ab - bc - ca) here,

a = 1/2, b = 1/3, c = −5/6 

a³ + b³ + c³ - 3abc

= (a + b + c)(a² + b² + c² - ab - bc - ca)

a³ + b³ + c³ = (a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (1/2 + 1/3 - 5/6)(a² + b² + c² - ab - bc - ca) + 3abc by

Using least common multiple

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (6/12 + 4/12 − 10/12)

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = 0

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = 3abc

(1/2)³ + (1/3)³ - (−5/6)³ = 3 × 1/2 × 1/3 × − 5/6 

= 1/2 * −5/6 = − 5/12 

Hence, the value of 

(1/2)³ + (1/3)³ − (5/6)³ is −5/12 

(d) (0.2)³ - (0.3)³ + (0.1)³ 

we know that, a³ + b³ + c³ - 3abc = (a + b + c)

(a² + b² + c² - ab - bc - ca) here,

a = 0.2, b = 0.3, c = 0.1

a³+ b³ + c³ - 3abc = (a + b + c)

(a² + b² + c² - ab - bc - ca) a³ + b³ + c³ = (a + b + c)

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (0.2 - 0.3 + 0.1)

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = (0)

(a² + b² + c² - ab - bc - ca) + 3abc

a³ + b³ + c³ = 3abc

(0.2)³ - (0.3)³ + (0.1)³ = 3abc

= 3(0.2)(- 0.3)(0.1) = – 0.018

Hence, the value (0.2)³ - (0.3)³ + (0.1)³ is 0.018