A CUBE MINUS B CUBE FORMULA

In this section, we are going to see the formula  for

a- b3

We already know the formula/expansion for (a - b)3.

That is, 

(a - b)3  =  a3 - b3 - 3ab(a - b)

Case 1 : 

(a - b)3  =  a3 - b3 - 3ab(a - b)

Add 3ab(a - b) to each side. 

(a - b)3 + 3ab(a - b)  =  a3 - b3

Therefore, the formula for (a3 - b3) is 

a3 - b3  =  (a - b)3 + 3ab(a - b)

Case 2 : 

From case 1, 

a3 - b3  =  (a - b)3 + 3ab(a - b)

a3 - b3  =  (a - b)[(a - b)2 + 3ab]

a3 - b3  =  (a - b)[a2 - 2ab + b+ 3ab]

a3 - b3  =  (a - b)(a2 + ab + b2)

Therefore, the formula for (a3 - b3) is 

a3 - b=  (a - b)(a2 + ab + b2)

So, 

(a - b) and (a2 + ab + b2)

are the factors of (a3 - b3).

Note : 

Based on our need, either we can use the formula in case 1 or in case 2 for (a3 - b3).

Practice Questions

Question 1 :

Factor :

x3 - 1

Solution :

Write (x3 - 1) in the form of (a3 - b3).

x3 - 1  =  x3 - 13

(x3 - 13) is in the form of (a3 - b3).

Comparing (a- b3) and (x- 13), we get

a  =  x

b  =  1

Write the formula for (a- b3) given in case 2 above.

a3 - b3  =  (a - b)(a2 + ab + b2)

Substitute x for a and 1 for b. 

x3 - 13  =  (x - 2)(x2 + x(1) + 12)

x3 + 1  =  (x - 1)(x2 + x + 1)

Question 2 :

Factor :

8x3 - 27y3

Solution :

Write (8x3 - 27y3) in the form of (a3 - b3).

8x3 - 27y3  =  (2x)3 - (3y)3

(2x)3 - (3y)3 is in the form of (a- b3).

Comparing (a- b3) and (2x)- (3y)3, we get

a  =  2x

b  =  3y

Write the formula for (a- b3) given in case 2 above.

a3 - b3  =  (a - b)(a2 + ab + b2)

Substitute 2x for a and 3y for b. 

(2x)3 - (3y)3  =  (2x - 3y)[(2x)2 + (2x)(3y) + (3y)2]

8x3 - 27y3  =  (2x - 3y)(4x2 + 6xy + 9y2)

Question 3 :

Factor :

125p3 - 64q3

Solution :

Write (125p3 - 64q3) in the form of (a3 - b3).

125p3 - 64q3  =  (5p)3 - (4q)3

(5p)3 - (4q)3 is in the form of (a- b3).

Comparing (a- b3) and (5p)- (4q)3, we get

a  =  5p

b  =  4q

Write the formula for (a- b3) given in case 2 above.

a3 - b3  =  (a - b)(a2 + ab + b2)

Substitute 5p for a and 4q for b. 

(5p)3 - (4q)3  =  (5p - 4q)[(5p)2 + (5p)(4q) + (4q)2]

125p3 - 64q3  =  (5p - 4q)(25p2 + 20pq + 16q2)

Question 4 :

Find the value of (m3 - n3), if m - n = 3 and mn = 28.

Solution :

Write (m- n3) in terms of (m - n) and mn using the formula given in case 1 above.

m3 - n3  =  (m - n)3 + 3mn(m - n)

Substitute 3 for (m - n) and 28 for xy. 

x3 - y3  =  (3)3 + 3(28)(3)

x3 - y3  =  27 + 252

x3 - y3  =  279

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