A CUBE MINUS B CUBE FORMULA

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

a³ - 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 + b² + 3ab]

a³ - b³  =  (a - b)(a² + ab + b²)

Therefore, the formula for (a³ - b³) is 

a³ - b³  =  (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 (a³ - b³) and (x³ - 1³), we get

a  =  x

b  =  1

Write the formula for (a³ - 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 (a³ - b³).

Comparing (a³ - b³) and (2x)³ - (3y)³, we get

a  =  2x

b  =  3y

Write the formula for (a³ - 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 (a³ - b³).

Comparing (a³ - b³) and (5p)³ - (4q)³, we get

a  =  5p

b  =  4q

Write the formula for (a³ - 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 (m³ - 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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