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# A Cube Plus B Cube Formula & A Cube Minus B Cube Formula

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Last updated date: 24th Jul 2024
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## Introduction

As we all know, maths contains formulas we need to learn for some of our chapters. The A cube B cube formula is one of the most common formulas of maths which is used in almost all chapters for basic solving and simplifying some equations. This comes under basic mathematics and needs to understand well.

In this article, we will understand the concept of a cube b cube formula, its expansion, the proof of this formula, and its types which are a cube plus b cube and a cube minus b cube.

## The A Cube B Cube Formula

The a cube b cube formula is defined as the sum or difference of cube of a and cube of b which are variables. These variables are used to easily put the values of variables and solve the given question.

The a cube b cube formula is in two variations, one is the addition, and the other is the subtraction - a cube plus b cube $\left( {{a}^{3}}+{{b}^{3}} \right)$ and a cube minus b cube $\left( {{a}^{3}}-{{b}^{3}} \right)$.

Now, let's understand how we can apply this formula to the given questions and which questions can be transformed into this formula.

## The $\left( {{a}^{3}}-{{b}^{3}} \right)$ Formula

The a cube minus b cube formula is given as,

$\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

## The $\left( {{a}^{3}}-{{b}^{3}} \right)$ Expansion

The $\left( {{a}^{3}}-{{b}^{3}} \right)$ formula can be expanded as,

$\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

On multiplying a and b separately,

$\left( {{a}^{3}}-{{b}^{3}} \right)=a\left( {{a}^{2}}+{{b}^{2}}+ab \right)-b\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

$=\left( {{a}^{3}}+a{{b}^{2}}+{{a}^{2}}b \right)-\left( b{{a}^{2}}+{{b}^{3}}+a{{b}^{2}} \right)$

As we can see, simple brackets are opened. The formula of a cube minus b cube is equal to the product of the difference between a and b and the sum of a square, an into b and b square.

## Formula Proof For $\left( {{a}^{3}}-{{b}^{3}} \right)$ Variations a3- b3

The proof of a cube minus b cube is explained in the below steps.

Step-1 To expand the cube minus b cube formula, solve the right-hand side of the formula.

Step-2 Now, first solve the brackets $\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

$=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

$=a\left( {{a}^{2}}+{{b}^{2}}+ab \right)-b\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ $={{a}^{3}}+a{{b}^{2}}+{{a}^{2}}b-b{{a}^{2}}-{{b}^{3}}-a{{b}^{2}}$

Step-3 On further solving and cancelling the terms, we get the respective formula.

${{a}^{3}}-{{b}^{3}}$

Thus, the left-hand side is equal to the right-hand side of the equation. Hence proved that a cube minus b cube is equal to the product of the difference between a and b and the sum of a square, a into b and b square.

## The $\left( {{a}^{3}}+{{b}^{3}} \right)$ Formula

The cube plus b cube formula is given as,

$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

## The $\left( {{a}^{3}}+{{b}^{3}} \right)$ Expansion

The $\left( {{a}^{3}}+{{b}^{3}} \right)$ formula can be expanded as,

$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

$=a\left( {{a}^{2}}+{{b}^{2}}-ab \right)+b\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

As we can see, simple brackets are opened. The formula of a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a into b.

## Formula Proof For $\left( {{a}^{3}}+{{b}^{3}} \right)$ Variations a3+ b3

The proof of a cube plus b cube is explained in the below steps.

Step-1 To expand the cube plus b cube formula, we are considering the right-hand side of the formula

Step-2 Now, first solve the brackets $\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

$\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

$=a\left( {{a}^{2}}+{{b}^{2}}-ab \right)+b\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

$={{a}^{3}}+a{{b}^{2}}-{{a}^{2}}b+b{{a}^{2}}+{{b}^{3}}-a{{b}^{2}}$

Step-3 On further solving and cancelling the terms, we get the respective formula.

${{a}^{3}}+{{b}^{3}}$

Thus, the left-hand side is equal to the right-hand side of the equation. Hence proved that a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a and b.

## Where to Apply a Cube B Cube Formula

To understand where to apply a cube b cube formula, observe the given question. For example,

For the direct questions, like, 63 - 23, we can simply apply the formula and solve, but for the questions like 8 - 53, we need to figure out if the given number is the cube of some other number. We can see that 8 is the cube of 2. Thus, we can write the given question as 23 - 53 and solve it.

This technique can also be used for a cube plus b cube formula.

This way, we can figure out where to apply the cube minus b cube and a cube plus b cube formula.

## Important Question

1. What will be the factor of 27x3+64?

Ans. Step 1 : write 27x3+64 in the form of $\left( {{a}^{3}}+{{b}^{3}} \right)$.

27 is the cube of 3. It can be written as :

$27{{x}^{3}}+64={{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}$

On comparing $\left( {{a}^{3}}+{{b}^{3}} \right)$ and ${{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}$, we get:

a=3x

b=4

According to the formula, $\left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$

${{\left( 3x \right)}^{3}}+{{\left( 4 \right)}^{3}}=\left( 3x+4 \right)\left( {{\left( 3x \right)}^{2}}+{{4}^{2}}+3x\times 4 \right)$

$27{{x}^{3}}+64=\left( 3x+4 \right)\left( 9{{x}^{2}}+16+12x \right)$

## Solved Problem

1. Solve x3 + y3 where values for x are 2 and y is 3.

Ans. We know the formula , $\left( {{x}^{3}}+{{y}^{3}} \right)=\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)$.

x=2 and y=3 (given)

$\left( {{2}^{3}}+{{3}^{3}} \right)=\left( 2+3 \right)\left( {{2}^{2}}+{{3}^{2}}-2\times 3 \right)$

$=\left( 2+3 \right)\left( 4+9-6 \right)$

$=35$

Therefore, the value for 23+33 is 35.

## Conclusion

The formula of a cube minus b cube is equal to the product of the difference between a and b and the sum of a square into b and b square. The formula of a cube plus b cube is equal to the product of the sum of a and b and the sum of a square and b square minus a into b.

## FAQs on A Cube Plus B Cube Formula & A Cube Minus B Cube Formula

1. Using the a3 - b3 formula, solve 43 - 63.

By using the formula , $\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.

a=4 and b=6  (given)

By substituting the values of a and b in the above formula,

${{4}^{3}}-{{6}^{3}}=\left( 4-6 \right)\left( {{4}^{2}}+{{6}^{2}}+4\times 6 \right)$

$=\left( 4-6 \right)\left( 16+36+24 \right)$

$=-152$

Therefore, 43-63= -152.

2. How can you solve a2- b2?

In order to solve we will use the right-hand side of the formula

a2- b2 $=\left( a+b \right)\left( a-b \right)$

$=a\left( a-b \right)+b\left( a-b \right)$

$={{a}^{2}}-ab+ba-{{b}^{2}}$

On cancelling the opposite sign values, we get

${{a}^{2}}-{{b}^{2}}$.

Therefore, the left-hand side is equal to the right-hand side.

3. Find the value of (x+y)2 when x = 4 and y = 3 .

Given : x = 4

Y = 3

We can solve this by using the formula :

${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$

${{\left( 4+3 \right)}^{2}}={{4}^{2}}+{{3}^{2}}+2\times 4\times 3$

$=16+9+24$

$=49$

Therefore, (4+3)2 = 49.