# a Cube minus b Cube formula [Proof , examples and practice problems]

In this post, you’re going to learn a very important formula a cube minus b cube with proof, examples, and practice problems. First, we will see proof of formula, then solve some examples based on this formula, and at the end, you will get practice problems with answers for practice.

To prove a³ – b³ formula ,you must know the expansion of (a -b)³

(a – b)³ = a³ – b³-3ab(a-b)

Now you may previously know formula of a³ – b³from your textbook .If not ,here are two expansion of a³ – b³ that you will find in your school textbook

1. a³ – b³= (a -b)³ +3ab(a-b)

2. a³ – b³= (a – b) + ( a² + b² +ab)

Let’s proof it one -by -one

Proof(1)

To prove: a³ – b³= (a -b)³ +3ab(a-b)

RHS= (a -b)³ +3ab(a-b)

We know expasion of (a -b)³ i.e (a – b)³ = a³ – b³-3ab(a-b)

RHS= a³ – b³-3ab(a-b) + 3ab(a-b)

RHS= a³ -b³ -3a²b + 3ab² + 3a²b -3ab²

RHS= a³ -b³-3a²b + 3ab² + 3a²b -3ab²

RHS= a³ -b³+ 3ab² 3ab²

RHS= a³ -b³

Hence proved

Let’s see another proof

Proof(2)

To prove: a³ – b³= (a – b) + ( a² + b² +ab)

Since ,we have already proved that,

a³ – b³= (a -b)³ +3ab(a-b)

Further , it can be written as

a³ – b³= [(a -b) (a-b)²] + 3ab(a-b)            [ Using Exponent law; (a -b)³=(a -b) (a-b)²]

Take (a -b) common from RHS

a³ – b³= (a -b) [ (a-b)² + 3ab)]

a³ – b³=(a -b) [ a² + b² -2ab + 3ab]

a³ – b³=(a -b) [ a² + b² + ab]

Hence proved

Let’s see some examples based on this concept

Examples

Example:01 If x – y = 4 and xy=21 ,find the value of x³ – y³

Solution:

We know ,

a³ – b³= (a -b)³ +3ab(a-b) , where a = x and b= y

∴ x³ – y³= (x -y)³ + 3xy(x-y)

Substitute x – y = 4 and xy=21

x³ – y³= (4)³ + 3(21)(4)

x³ – y³= 64 + 252

x³ – y³= 316

Example:02 If a – b = 4 and ab=45 ,find the value of a³- b³

Solution:

This is the same type of problem, as we solved above. But, let’s solve it with the second approach

We know,

a³ – b³=(a -b) [ a² + b² + ab]

We have a – b = 4 and ab=45 but we don’t have value of a² + b² .So, to solve this simply convert a² + b² into term of (a -b) .How ? see

a³ – b³=(a -b) [ a² + b² + ab]

a³ – b³=(a -b) [ a² – 2ab + b² + 2ab +ab]  (Add and subtract 2ab )

a³ – b³=(a -b)[(a -b)²+ 3ab]

Substitute a – b = 4 and ab=45

a³ – b³= (4)[(4)² + 3(45)]

a³ – b³=4( 16 + 135)

a³ – b³=4(151)

a³ – b³= 604

Example:03 Find the products

( 7a – 5b) ( 49a² + 35ab + 25b²)

Solution:

=( 7a – 5b) [(7a)² + 7a × 5b + (5b)²]

If you see it carefully ,it is of form

=(a -b)(a² + ab +b²) where a = 7a and b = 5b

= a³ – b³                                     [ a³ – b³=(a -b) [ a² + b² + ab]

Substitute a = 7a and b= 5b

= (7a)³ – (5b)³

= 343a³ – 125b³

Example:04  If a – b = 10 and ab=7 ,find the value of (a²+ ab +b²)

Solution:

We know ,

a³ – b³ = (a – b) (a²+ ab +b²)

To get the value of (a²+ ab +b²) from above equation ,we require value of a³ – b³.

Second expansion of a³ – b³

a³ – b³= (a -b)³ +3ab(a-b)

Susbtitute the value of a – b = 10 and ab=7

a³ – b³= (a -b)³ +3ab(a-b)

a³ – b³=(10)³ + 3(7)(10)

a³ – b³=1000 + 210

a³ – b³=1210

Now ,we can easily find the value of (a²+ ab +b²)

a³ – b³ = (a – b) (a²+ ab +b²)

1210 = (10) (a²+ ab +b²)

(a²+ ab +b²) = 121

Practice problem

Q.1 If a – b = 4 and ab= 21 ,find the value of a³ – b³

Q.2 (x² -1)(x^4 +x²+1)

Q.3If a -b = 6 and ab=20 ,find the value of a³-b³