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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » A Cube Plus B Cube

A Cube Plus B Cube

In this article, we are going to understand the formula a3+b3 with examples, and also understand the concept of Algebra.

Table of Content
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The sum of cubes (of two numbers) formula is represented as a3 + b3. Without having to calculate the cubes, the a cube plus b cube formula is used to obtain the sum of the two cubes. It is also employed in the factorization of cube binomials. In this section, we’ll go through the many aspects of the a3 + b3 formula.

The sum of cubes of the two numbers is also known as the formula of a cube plus b cube. This algebraic formula can be used to compute the sum of cubes of two numbers. A binomial is a cube + b cube (a3 + b3) that can be factored using the formula.

You may evaluate the a3 + b3 formula by multiplying (a + b) (a2 – ab + b2) and seeing if you obtain a3 + b3. The a3 + b3 formula, sometimes known as the difference of cubes formula, is as follows:

a3 + b3 = (a + b) (a2 – ab + b2) 

Proof of a3 + b3 Formula

Let us see the proof of a cube plus b cube a3 + b3 formula below. 

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

we need to prove here LHS = RHS. Let’s begin with the following steps.

LHS = a3 + b3

On Solving RHS side we get,

= (a + b) (a2 – ab + b2)

On multiplying the a and b separately with (a2 – ab + b2) we get

= a (a2 – ab + b2) + b(a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3

= a3 – a2b + a2b + ab2– ab2 + b3

= a3 – 0 + 0 + b3

= a3 + b3

Hence proved, LHS = RHS

Now Let us see the proof of a cube plus b cube a3 + b3 formula below. 

To prove that (a + b)3  =  (a3 + b3 )- 3ab(a + b)

we need to prove here LHS = RHS. Let’s begin with the below steps.

(a + b)3  =  (a3 + b3 )- 3ab(a + b)

Now, Subtract the 3ab (a + b) from each side of the above Equation. 

(a + b)3 – 3ab(a + b)  =  a3 + b3

Therefore, the formula for (a3 + b3) is 

(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 + b3)=  (a + b)(a2 – ab + b2)

So, 

(a + b) and (a2 – ab + b2)

are the factors of (a3 + b3)

Thus, (a + b)3  =  (a3 + b3 )- 3ab(a + b)

How to Use a3+b3 Formula?

The following steps can be taken while using a3 + b3 formula.

  • Firstly, observe the pattern of the 2 numbers and check the numbers have 3 as power or not.
  • After that, note down the formula of and that is a3 + b3 = ( a  +  b) ( a 2 –  a b + b2)
  • Now, Substitute the values of the a and the b in a3 + b3 formula.
  • Lastly, multiply a and b one by one with (a2 – ab + b2) and simplify to get the result.

Point To remember

  • (a + b)3  =  (a3 + b3 )- 3ab(a + b)
  • a3 + b3 = (a + b) (a2 – ab + b2)
faq

Frequently asked questions

Get answers to the most common queries related to the JEE Examination Preparation.

Write Down the formula of (a3 + b3)?

Ans : The Formula of (a3...Read full

Find value of 1003 + 23 by using the a3 + b3 = (a + b) (a2 - ab + b2) formula.

Ans :  We know that,a3...Read full

Write down the step to use (a3 + b3)?

Ans : The step to use (a3...Read full

Write down the proof of (a + b)3 = (a3 + b3 )- 3ab(a + b)?

Ans: A/Q we have to Prove the Algebraic Formula (a + b)3...Read full

Ans : The Formula of (a3 + b3) are:

  • (a + b)(a2 – ab + b2)
  • {a3 + b3) – 3ab(a + b)

 

Ans :  We know that,a3 + b3 = {(a + b) (a2 – ab + b2) Here a=100 and b=2

Let us put this value we get,

1003 + 23 = (100 + 2) (1002 – 100*2 + 22) 

1003 + 23 = 1000008

 

Ans : The step to use (a3 + b3) are below:

  • Step 1: Start with observing the pattern of the 2 numbers and check the numbers have 3 as power or not.
  • Step 2: note down the formula a3 + b3 = ( a  +  b) ( a 2 –  a b + b2)
  • Step 3: Substitute the values of the a and the b in a3 + b3 formula.
  • Step 4: multiply a and b one by one with (a2 – ab + b2) and simplify to get the result.

 

Ans: A/Q we have to Prove the Algebraic Formula (a + b)3  =  (a3 + b3 )- 3ab(a + b)

to prove Lets take LHS = RHS. 

Now begin with the Following steps:

(a + b)3  =  (a3 + b3 )- 3ab(a + b)

Now, Subtract the 3ab(a + b) from each side of the above Equation. 

(a + b)3 – 3ab(a + b)  =  a3 + b3

Therefore, the formula for (a3 + b3) is 

(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 + b3)=  (a + b)(a2 – ab + b2)

So, 

(a + b) and (a2 – ab + b2)

are the factors of (a3 + b3)

Thus, (a + b)3  =  (a3 + b3 )- 3ab(a + b)

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