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)³ 

Solution :

(x + 2)³ is in the form of (a + b)³

Comparing (a + b)³ 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)³ 

Solution :

(2x + 3)³ is in the form of (a + b)³

Comparing (a + b)³ 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)³ 

Solution :

(p + 2q)³ is in the form of (a + b)³

Comparing (a + b)³ 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)³ and find the value of (107)³ easily.

Write (107)³ 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, 

Please click here

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