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