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# 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)

⟹ 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)

⟹ 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,

⟹ (x- 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 – 729x... 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 - 64y+ 125z3 + 180xyz

Hence, the value of (3x - 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy - 15zx + 20yz) is 27x3 - 64y+ 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)= x2 + y2 + z2 + 2(20)

(x + y + z)= x2 + y2 + z2+ 40

82 = x2 + y2 + z2 + 40

64 - 40 = x2 + y2 + z2

x2 + y2 + z= 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 + z= 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)= a2 + b2 + c2 + 2(26)

(a + b + c)= a2 + b2+ c2 + 52

92 = a2 + b2 + c2 + 52

81 - 52 = a2 + b2 + c2

a2 + b2 + c= 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 + c= 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)is 0.018