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 + y= (x + y)3 - 3xy(x + y)2

x3 - y= (x - y)3 + 3xy(x - y)2

Example 1 :

Expand (x - y)3.

Solution :

We know that

(x + y)= 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)= x3 - 3xy(x - y) - y3

Example 2 :

Expand (x + 4)3.

Solution :

We know that

(x + y)= 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)= 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)= 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)= 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)= x3 + 3x2y + 3xy2 + y3

Substitute x = 100, and y = -2.

(100 - 2)3 = 1003 + 3(100)2(-2) + 3(100)(-2)2 + (-2)3

98= 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)= 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 + y= (x + y)3 - 3xy(x + y)

Substitute x = 3a, and y = 4b.

(3a)3 + (4b)= (3a + 4b)3 - 3(3a)(4b)(3a + 4b)

27a3 + 64b= (3a + 4b)3 - 36(ab)(3a + 4b)

Substitute 3a + 4b = 10 and ab = 2.

27a3 + 64b= (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 - y= (x - y)3 + 3xy(x - y)

Substitute x = p, and y = q.

p3 - q= (p - q)3 + 3pq(p - q)

Substitute p - q = 5 and pq = 14.

p3 - q= 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 + y= (x + y)3 - 3xy(x + y)

Substitute x = x, and y = 1/x.

x3 + (1/x)= (x + 1/x)3 - 3(x)(1/x)(x + 1/x)

x3 + 1/x= (x + 1/x)3 - 3(x + 1/x)

Substitute x + 1/x = 6.

x3 + 1/x= (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 - y= (x - y)3 + 3xy(x - y)

Substitute x = b, and y = 1/b.

b3 - (1/b)= (b - 1/b)3 + 3(b)(1/b)(b - 1/b)

b3 - 1/b= (b - 1/b)3 + 3(b - 1/b)

Substitute (b - 1/b)3 = 27 and b - 1/b = 3.

b3 - 1/b= 27 + 3(3)

= 27 + 9

= 36

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