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) 

Start with RHS 

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

Answer: 316

Q.2 (x² -1)(x^4 +x²+1) 

Answer: x^6 – 1

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

Answer: 576

Q.4 Find the value of 64x³ – 125z³ ,if 4x – 5z = 16 and xz=12 .

Answer: 15616

Check these resources as well 

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