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 is 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)

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 (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 the 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

That’s it from our side

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