a3 b3 Formula or a^{3} b^{3} 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 is as follows.

a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

a^{3} – b^{3} = (a – b)(a^{2} + ab +b^{2})

**Proof of a3 + b3 formula**

As we know that:

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Hence:

a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)

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

a^{3} + b^{3} = (a + b)[(a + b)^{2} – 3ab]

a^{3} + b^{3} = (a + b)[(a^{2} + b^{2} + 2ab) – 3ab]

a^{3} + b^{3} = (a + b)(a^{2} + b^{2} + 2ab –3ab)

a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

**Proof of a3 – b3 formula**

As we know that:

(a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

Hence:

a^{3} – b^{3} = (a – b)^{3} + 3ab(a – b)

Taking (a – b) common in RHS, we get

a^{3} – b^{3} = (a – b)[(a – b)^{2} + 3ab]

a^{3} – b^{3} = (a – b)[(a^{2} + b^{2} – 2ab) + 3ab]

a^{3} – b^{3} = (a – b)(a^{2} + b^{2} – 2ab + 3ab)

a^{3} – b^{3} = (a – b)(a^{2} + b^{2} + ab)

See Also: a2 b2 formula

**Solved Questions and Answers for a3 b3 Formula**

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

**Q1) If a = 7 and b = 3,**

**Find the value of (a ^{2} – ab + b^{2}) using a3 b3 Formula**

Ans)

Given: a = 7 and b = 3

As per the given data,

a + b = 10

And

a^{3} + b^{3} = 7^{3} + 3^{3}

a^{3} + b^{3} = 343 + 27

a^{3} + b^{3} = 370

As we know that

a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

Substituting value of a^{3} + b^{3} and a + b:

370 = 10 X (a^{2} – ab + b^{2})

Hence

(a^{2} – ab + b^{2}) = 370/10

(a^{2} – ab + b^{2}) = 37

#### Practice questions

**Ques 1) If a = 7 and ab = 28**

**Find the value of a ^{3} – b^{3} using a3 b3 Formula**

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

**Find the value of a ^{3} + b^{3} using a3 b3 Formula**

## a3 b3 c3 Formula

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)

**Proof of a3 b3 c3 Formula**

The following is the proof of a3 b3 c3 formula

We take LHS

a^{3} + b^{3} + c^{3} – 3abc

Adding and subtracting 3ab(a + b)

a^{3} + b^{3} + 3ab(a + b) – 3ab(a + b) + c^{3} – 3abc

As we know that: (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

[a^{3} + b^{3} + 3ab(a + b)] – 3ab(a + b) + c^{3} – 3abc

(a + b)^{3 }– 3ab(a + b) + c^{3} – 3abc

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

(a + b)^{3 }+ c^{3} + 3(a + b)c(a + b + c) – 3(a + b)c(a + b + c) – 3ab(a + b) – 3abc

[(a + b)^{3 }+ c^{3} + 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} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

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

(a + b + c)[a^{2} + b^{2} + c^{2} – (ab + bc + ca)]

(a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

Hence Proved

### Solved questions using a3 b3 c3 Formula

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

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

Find the value of (a^{2} + b^{2} + c^{2} – ab – bc – ca) using a3 b3 c3 Formula

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

Hence,

a + b + c = 12

a^{3} = 4^{3} = 64

b^{3} = 7^{3} = 343

c^{3}= 1^{3} = 1

abc = 4 X 7 X 1 = 28

a^{3} + b^{3} + c^{3 }= 408

Substituting the values in the equation

a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)

408 – 3 X 28 = 12 X (a^{2} + b^{2} + c^{2} – ab – bc – ca)

324 = 12 X (a^{2} + b^{2} + c^{2} – ab – bc – ca)

Hence,

(a^{2} + b^{2} + c^{2} – ab – bc – ca) = 324/12

(a^{2} + b^{2} + c^{2} – ab – bc – ca) = 27

#### Practice questions

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

**Find the value of [a ^{3} + b^{3} + c^{3}– 3abc] using a3 b3 c3 formula**

That’s it from our side

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