## a plus b WHOLE CUBE FORMULA

In this section, you will learn the formula or expansion  for (a + b)3.

That is,

(a + b)3  =  (a + b)(a + b)(a + b)

Multiply (a + b) and (a + b).

(a + b)3  =  (a2 + ab + ab + b2)(a + b)

Simplify.

(a + b)3  =  (a2 + 2ab + b2)(a + b)

(a + b)3  =  a3 + a2b + 2a2b + 2ab2 + ab2 + b3

Combine the like terms.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

or

(a + b)3  =  a3 + b3 + 3ab(a + b)

## Solved Problems

Problem 1 :

Expand :

(x + 2)

Solution :

(x + 2)is in the form of (a + b)3

Comparing (a + b)and (x + 2)3, we get

a  =  x

b  =  2

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute x for a and 2 for b.

(x + 2)3  =  x3 + 3(x2)(2) + 3(x)(22) + 23

(x + 2)3  =  x3 + 6x2 + 3(x)(4) + 8

(x + 2)3  =  x3 + 6x2 + 12x + 8

So, the expansion of (x + 2)3 is

x3 + 6x2 + 12x + 8

Problem 2 :

Expand :

(2x + 3)

Solution :

(2x + 3)is in the form of (a + b)3

Comparing (a + b)and (2x + 3)3, we get

a  =  2x

b  =  3

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute 2x for a and 3 for b.

(2x + 3)3  =  (2x)3 + 3(2x)2(3) + 3(2x)(32) + 33

(2x + 3)3  =  8x3 + 3(4x2)(3) + 3(2x)(9) + 27

(2x + 3)3  =  8x3 + 36x2 + 54x + 27

So, the expansion of (2x + 3)3 is

8x3 + 36x2 + 54x + 27

Problem 3 :

Expand :

(p + 2q)

Solution :

(p + 2q)is in the form of (a + b)3

Comparing (a + b)and (p + 2q)3, we get

a  =  p

b  =  2q

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

Substitute p for a and 2q for b.

(p + 2q)3  =  p3 + 3(p2)(2q) + 3(p)(2q)2 + (2q)3

(p + 2q)3  =  p3 + 6p2q + 3(p)(4q2) + 8q3

(p + 2q)3  =  p3 + 6p2q + 12pq2 + 8q3

So, the expansion of (p + 2q)3 is

p3 + 6p2q + 12pq2 + 8q3

Problem 4 :

If a + b  =  12 and a3 + b3  =  468, then find the value of ab.

Solution :

To find the value of ab, we can use the formula or expansion for (a + b)3.

Write the formula / expansion for (a + b)3.

(a + b)3  =  a3 + 3a2b + 3ab2 + b3

or

(a + b)3  =  a3 + b3 + 3ab(a + b)

Substitute 12 for (a + b) and 468 for (a3 + b3).

(12)3  =  468 + 3(ab)(12)

Simplify.

1728  =  468 + 36ab

Subtract 468 from each side.

1260 =  36ab

Divide each side by 36.

35  =  ab

So, the value of ab is 35.

Problem 5 :

Find the value of :

(107)3

Solution :

We can use the algebraic formula for (a + b)and find the value of (107)easily.

Write (107)in the form of (a + b)3.

(107)3  =  (100 + 7)3

Write the expansion for (a + b)3.

(a + b)3  =  a3 + b3 + 3ab(a + b)

Substitute 100 for a and 7 for b.

(100 + 7)3  =  1003 + 73 + 3(100)(7)(100 + 7)

(100 + 7)3  =  1000000 + 343 + 3(100)(7)(107)

(100 + 7)3  =  1000000 + 343 + 224700

(107)3  =  1225043

So, the value of (107)3 is

1,225,043 Apart from the stuff explained above, if you would like to learn about more identities in Algebra,

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