A CUBE PLUS B CUBE FORMULA

In this section, you will learn the formula or expansion 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)

Subtract 3ab(a + b) from 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).

Solved Questions

Question 1 :

Factor :

x³ + 8

Solution :

Write (x³ + 8) in the form of (a³ + b³).

x³ + 8  =  x³ + 2³

(x³ + 2³) is in the form of (a³ + b³).

Comparing (a³ + b³) and (x³ + 2³), we get

a  =  x

b  =  2

Write the formula for (a³ + b³) given in case 2 above.

a³ + b³  =  (a + b)(a² - ab + b²)

Substitute x for a and 2 for b. 

x³ + 2³  =  (x + 2)(x² - 2x + 2²)

x³ + 8  =  (x + 2)(x² - 2x + 4)

Question 2 :

Factor :

27x³ + 64

Solution :

Write (27x³ + 64) in the form of (a³ + b³).

27x³ + 64  =  (3x)³ + 4³

(3x)³ + 4³ is in the form of (a³ + b³).

Comparing (a³ + b³) and (3x)³ + 4³, we get

a  =  3x

b  =  4

Write the formula for (a³ + b³) given in case 2 above.

a³ + b³  =  (a + b)(a² - ab + b²)

Substitute 3x for a and 4 for b. 

(3x)³ + 4³  =  (3x + 4)[(3x)² - (3x)(4) + 4²]

27x³ + 64  =  (3x + 4)(9x² - 12x + 16)

Question 3 :

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)(8x² - 6xy + 9y²)

Question 4 :

Find the value of (x³ + y³), if x + y = 4 and xy = 5.

Solution :

Write (x³ + y³) in terms of (x + y) and xy using the formula given in case 1 above.

x³ + y³  =  (x + y)³ - 3xy(x + y)

Substitute 4 for (x + y) and 5 for xy. 

x³ + y³  =  (4)³ - 3(5)(4)

x³ + y³  =  64 - 60

x³ + y³  =  4

Question 5 :

Find the value of

(i) x³ + y³ – 12xy + 64, when x + y = – 4

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6

Solution :

(i) x³ + y³ – 12xy + 64, when x + y = – 4

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

x³ + y³ + 4³ – 12xy

= (x + y + 4)(x² + y² + 4² - x(y) - 4(y) - 4x)

= (x + y + 4)(x² + y² + 16 - xy - 4y - 4x)

Applying x + y = -4, we get

= (-4 + 4)(x² + y² + 16 - xy - 4y - 4x)

x³ + y³ – 12xy + 64 = 0

So, the required value is 0.

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6

x³ + (-2y)³ + (-6)³ – 36xy = (x - 2y - 6)(x² + 4y² + 6² - x(-2y) - (-2y)(-6) - (-6)(-2y)

Applying x = 2y + 6, we get

= (2y + 6 - 2y - 6)(x² + 4y² + 6² - x(-2y) - (-2y)(-6) - (-6)(-2y)

x³ – 8y³ – 36xy – 216 = 0

So, the required result is 0.

Question 6 :

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(n³ - 3n²m + 3nm² - m³) / (5m² - 10mn + 5n²)

Solution :

(n³ - 3n²m + 3nm² - m³) / (5m² - 10mn + 5n²)

(n³ - 3n²m + 3nm² - m³) = (n - m)³ ------(1)

(5m² - 10mn + 5n²) = (5m² - 5mn - 5mn + 5n²)

= 5m(m - n) - 5n(m - n)

= (m - n) 5(m - n)

= 5(m - n)² ------(2)

(1) / (2)

= (n - m)³ /  5(m - n)² 

= (n - m) / 5

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