## A CUBE PLUS B CUBE FORMULA

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

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 + 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).

## 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 (a+ b3) and (x+ 23), we get

a  =  x

b  =  2

Write the formula for (a+ 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 (a+ b3).

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

a  =  3x

b  =  4

Write the formula for (a+ 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 (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)(8x2 - 6xy + 9y2)

Question 4 :

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

Solution :

Write (x+ 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 Apart from the stuff given above, if you need any other stuff, please use our google custom search here.

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