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

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