CUBE FORMULAS IN ALGEBRA
Expansion of (x + y)3 :
(x + y)3 = (x + y)(x + y)(x + y)
= [(x + y)(x + y)](x + y)
= [x2 + xy + xy + y2](x + y)
= [x2 + 2xy + y2](x + y)
= x2(x) + 2xy(x) + y2(x) + x2(y) + 2xy(y) + y2(y)
= x3 + 2x2y + xy2 + x2y + 2xy2 + y3
= x3 + 3x2y + 3xy2 + y3
or
= a3 + 3aby(a + b) + b3
Thus,
(x + y)3 = x3 + 3xy(x + y) + y3
or
(x + y)3 = x3 + 3xy(x + y) + y3
Identities involving sum, difference and product are stated here :
x3 + y3 = (x + y)3 - 3xy(x + y)2
x3 - y3 = (x - y)3 + 3xy(x - y)2
Example 1 :
Expand (x - y)3.
Solution :
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Replace 'y' by '-y'.
(x + (-y))3 = x3 + 3x2(-y) + 3xx(-y)2 + (-y)3
(x - y)3 = x3 - 3x2y + 3xy2 - y3
or
(x - y)3 = x3 - 3xy(x - y) - y3
Example 2 :
Expand (x + 4)3.
Solution :
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute x = x, and y = 4.
(x + 4)3 = x3 + 3(x)2(4) + 3(x)(4)2 + 43
= x3 + 12x2 + 3(x)(16) + 64
= x3 + 12x2 + 48x + 64
Example 2 :
Expand (5a + 3b)3.
Solution :
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute x = 5a, and y = 3b.
(5a + 3b)3 = (5a)3 + 3(5a)2(3b) + 3(5a)(3b)2 + (3b)3
= 125a3 + 3(25a2)(3b) + 3(5a)(9b2) + 27b3
= 125a3 + 225a2b + 135ab2 + 27b3
Example 3 :
Expand (3x - 4y)3.
Solution :
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute x = 3x, and y = -4y.
(3x - 4y)3 = (3x)3 + 3(3x)2(-4y) + 3(3p)(-4q)2 + (-4y)3
= 27x3 + 3(9x2)(-4y) + 3(3x)(16y2) + (-64y3)
= 27x3 - 108x2y + 144xy2 - 64y3
Example 4 :
Expand (a + 1/b)3.
Solution :
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute a = x, and b = 1/y.
(x + 1/y)3 = x3 + 3(x)2(1/y) + 3(x)(1/y)2 + (1/y)3
= x3 + 3x2/y + 3x/y2 + 1/y3
Example 5 :
Evaluate using identity : 983.
Solution :
983 = (100 - 2)3
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute x = 100, and y = -2.
(100 - 2)3 = 1003 + 3(100)2(-2) + 3(100)(-2)2 + (-2)3
983 = 1000000 + 3(10000)(-2) + 3(100)(4) - 8
= 1000000 - 60000 + 1200 - 8
= 941192
Example 6 :
Evaluate using identity : 10013.
Solution :
10013 = (1000 + 1)3
We know that
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Substitute x = 1000, and y = 1.
(1000 + 1)3 = 10003 + 3(1000)2(1) + 3(1000)(1)2 + (1)3
(1001)3 = 10003 + 3(1000)2(1) + 3(1000)(1)2 + (1)3
= 1000000000 + 3(1000000)(1) + 3(1000)(1) + 1
= 1000000000 + 3000000 + 3000 + 1
= 1003003001
Example 7 :
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2.
Solution :
We know that
x3 + y3 = (x + y)3 - 3xy(x + y)
Substitute x = 3a, and y = 4b.
(3a)3 + (4b)3 = (3a + 4b)3 - 3(3a)(4b)(3a + 4b)
27a3 + 64b3 = (3a + 4b)3 - 36(ab)(3a + 4b)
Substitute 3a + 4b = 10 and ab = 2.
27a3 + 64b3 = (10)3 - 36(2)(10)
= 1000 - 720
= 280
Example 8 :
Find p3 - q3, if p - q = 5 and pq = 14.
Solution :
We know that
x3 - y3 = (x - y)3 + 3xy(x - y)
Substitute x = p, and y = q.
p3 - q3 = (p - q)3 + 3pq(p - q)
Substitute p - q = 5 and pq = 14.
p3 - q3 = 53 + 3(14)(5)
= 125 + 210
= 335
Example 9 :
If x + 1/x = 6, then find the value of x3 + 1/x3.
Solution :
We know that
x3 + y3 = (x + y)3 - 3xy(x + y)
Substitute x = x, and y = 1/x.
x3 + (1/x)3 = (x + 1/x)3 - 3(x)(1/x)(x + 1/x)
x3 + 1/x3 = (x + 1/x)3 - 3(x + 1/x)
Substitute x + 1/x = 6.
x3 + 1/x3 = (6)3 + 3(6)
= 216 + 18
= 198
Example 10 :
If (b - 1/b)3 = 27, then find the value of b3 - 1/b3.
Solution :
(b - 1/b)3 = 27
(b - 1/b)3 = 33
b - 1/b = 3
We know that
x3 - y3 = (x - y)3 + 3xy(x - y)
Substitute x = b, and y = 1/b.
b3 - (1/b)3 = (b - 1/b)3 + 3(b)(1/b)(b - 1/b)
b3 - 1/b3 = (b - 1/b)3 + 3(b - 1/b)
Substitute (b - 1/b)3 = 27 and b - 1/b = 3.
b3 - 1/b3 = 27 + 3(3)
= 27 + 9
= 36
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