# a3 b3 Formula – with Questions and Answers

a3 b3 Formula or a3 b3 Formula is one of the most important and most asked formulae in Mathematics. Understanding the formula and applying it to solve the questions is easy. In this article, we explain the a3 b3 formula and a3 b3 c3 formula with proofs, along with solved questions, as well as unsolved for your practice.

## a3 b3 Formula and a3 b3 Formula Proof

a3 b3 Formula is a polynomial equation with two variables. It is an equation in the form P=0, where P is the polynomial. A polynomial is a mathematical expression that solely uses the operations of subtraction, addition, multiplication, and positive-integer powers of variables. It is made up of indeterminates, also known as variables, such as x, y and z, and numeric coefficients.

### The a3 b3 Formula are as follows.

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab +b2)

Proof of a3 + b3 formula

As we know that:

(a + b)3 = a3 + b3 + 3ab(a + b)

Hence:

a3 + b3 = (a + b)3 – 3ab(a + b)

Taking (a + b) common in RHS, we get

a3 + b3 = (a + b)[(a + b)2 – 3ab]

a3 + b3 = (a + b)[(a2 + b2 + 2ab) – 3ab]

a3 + b3 = (a + b)(a2 + b2 + 2ab –3ab)

a3 + b3 = (a + b)(a2 – ab + b2)

Proof of a3 – b3 formula

As we know that:

(a – b)3 = a3 – b3 – 3ab(a – b)

Hence:

a3 – b3 = (a – b)3 + 3ab(a – b)

Taking (a – b) common in RHS, we get

a3 – b3 = (a – b)[(a – b)2 + 3ab]

a3 – b3 = (a – b)[(a2 + b2 – 2ab) + 3ab]

a3 – b3 = (a – b)(a2 + b2 – 2ab + 3ab)

a3 – b3 = (a – b)(a2 + b2 + ab)

### Solved Questions and Answers for a3 b3 Formula

In this section, we have provided a few solved questions that involve usage of a3 b3 Formula.

Q1) If a = 7 and b = 3,

Find the value of (a2 – ab + b2) using a3 b3 Formula

Ans)

Given: a = 7 and b = 3

As per the given data,

a + b = 10

And

a3 + b3 = 73 + 33

a3 + b3 = 343 + 27

a3 + b3 = 370

As we know that

a3 + b3 = (a + b)(a2 – ab + b2)

Substituting value of a3 + b3 and a + b:

370 = 10 X (a2 – ab + b2)

Hence

(a2 – ab + b2) = 370/10

(a2 – ab + b2) = 37

#### Practice questions

Ques 1) If a = 7 and ab = 28

Find the value of a3 – b3 using a3 b3 Formula

Ques 2) If a + b = 11 and ab = 18

Find the value of a3 + b3 using a3 b3 Formula

## a3 b3 c3 Formula

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

Proof of a3 b3 c3 Formula

The following is the proof of a3 b3 c3 formula

We take LHS

a3 + b3 + c3 – 3abc

Adding and subtracting 3ab(a + b)

a3 + b3 + 3ab(a + b) – 3ab(a + b) + c3 – 3abc

As we know that: (a + b)3 = a3 + b3 + 3ab(a + b)

[a3 + b3 + 3ab(a + b)] – 3ab(a + b) + c3 – 3abc

(a + b)3 – 3ab(a + b) + c3 – 3abc

Adding and subtracting 3(a + b)c(a + b + c)

(a + b)3 + c3 + 3(a + b)c(a + b + c) – 3(a + b)c(a + b + c) – 3ab(a + b) – 3abc

[(a + b)3 + c3 + 3(a + b)c(a + b + c)] – 3(a + b)c(a + b + c) – 3ab(a + b + c)

(a + b + c)3  – 3(a + b)c(a + b + c) – 3ab(a + b + c)

Taking (a + b + c) common

(a + b + c)[(a + b + c)2  – 3(a + b)c – 3ab]

(a + b + c)[(a + b + c)2 – 3(ab + bc + ca)]

As we know that: (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

(a + b + c)[a2 + b2 + c2 + 2(ab + bc + ca) – 3(ab + bc + ca)]

(a + b + c)[a2 + b2 + c2 – (ab + bc + ca)]

(a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Hence Proved

### Solved questions using a3 b3 c3 Formula

In this section, we have provided a few solved questions that involve usage of a3 b3 c3 Formula.

Ques) If a = 4, b = 7 and c = 1

Find the value of (a2 + b2 + c2 – ab – bc – ca) using a3 b3 c3 Formula

Ans) Given: a = 4, b = 7 and c = 1

Hence,

a + b + c = 12

a3 = 43 = 64

b3 = 73 = 343

c3= 13 = 1

abc = 4 X 7 X 1 = 28

a3 + b3 + c3 = 408

Substituting the values in the equation

a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

408 – 3 X 28 = 12 X (a2 + b2 + c2 – ab – bc – ca)

324 = 12 X (a2 + b2 + c2 – ab – bc – ca)

Hence,

(a2 + b2 + c2 – ab – bc – ca) = 324/12

(a2 + b2 + c2 – ab – bc – ca) = 27

#### Practice questions

Ques) If a = 7, b = 8, c = 9

Find the value of [a3 + b3 + c3– 3abc] using a3 b3 c3 formula

Thats it from our side