Algebra is a mathematical concept that helps people visualise difficult situations by utilising mathematical equations using variables such as x, y, and z. Here we will explain to you the algebraic formula of a³b³ with its proof derivative and some examples. This formula has more weightage not only for school students but also for the students preparing for competitive examinations like NTSE, NDA, AFCAT, SSC, Railways, etc.
a3b3 Formula
The formula of a3b3 in algebra is a³b³= (ab) (a²+ab+b²). It is also known as a cube minus b cube formula. You can use the a3b3 formula in maths to solve different algebraic problems.
a³b³ = (a – b) (a²+ab+b²)
Basics of Algebraic Equation
The Algebraic Equation in the subject of mathematics is a statement where two algebraic expressions are set equal to one another. The combination of variables, coefficients, and constants makes up the Algebraic Equation.
An Algebraic Equation always provides the balanced expression of unknown variables, constants, and coefficients. As both sides of the equation have the same value, it is considered a balanced equation. It is presented in the form of P=0 where P refers to Polynomial.
a3 b3 Formula Chart
There are many other formulas related to a³ b³ in algebra. Let’s have a look at the a³ b³ Formula Chart.
a3 b3 Formula Chart (Algebra Formula) 

a3b3 Formula Proof
There are various algebraic formulas in mathematics to be used for solving the equation. Here the complete proof derivation of the a3b3 formula is given below.
Since we know the formula of (ab)³= a³b³3ab(ab)
a³b³= (ab)³+ 3ab(ab)
a³b³= (ab) [(ab)² + 3ab]
After the expansion of the equation on Right Hand Side (RHS),
a3b³= (a – b) (a²+b²2ab+3ab)
a3b³= (a – b) (a²+b²+ab)
a3b³= (a – b) (a²+ab+b²)
The above equation can be proved by the following steps:
First considering the Right Hand Side (RHS) of the equation,
(ab) (a²+ ab+ b²) = a(a² + ab + b²) – b(a² + ab + b²)
(ab) (a²+ ab+ b²)= (a – b)(a² + ab + b²) = a3 + a2b + ab2 – a2b – ab2 – b3
Taking out the common factor (ab),
(ab) (a2+ ab+ b2)= a3 + a2b – a2b + ab2– ab2 – b3
Bringing the similar terms together and canceling out the probable terms such as a2b and ab2,
(ab) (a2+ ab+ b2)= a3b3 (RHS) which is equal to Left Hand Side (LHS). Hence this formula is proved.
In this way, you should remember the a³b³ formula to start your calculation in the examination.
a³b³ = (a – b) (a²+ab+b²)
a³b³ Formula Examples with Solutions
By memorizing this formula, you can easily calculate the mathematical problems.
Question 1: Factorize 125a³27b³?
Solution: As we know that the expression of 125a³27b³ can be written as (5a)³(3b)³
By employing the a³b³ formula,
a³b³ = (a – b) (a²+ab+b²)
Now, we will get the factors of the given equation as,
(5a)³(3b)³ = (5a3b) (25a² + 15ab + 9b²)
Here 5a indicates a and 3b indicates b of the wellproven a³b³ formula.
Question 2: Factorize (3a + b)³ – (2a + b)³?
Solution: As we know that the given equation is in the form of the a3b3 formula,
a³ – b³= (3a + b)³ – (2a + b)³
Here a is indicated by (3a + b) while b is indicated by (2a + b)
By using the formula of a³b³,
a3b3= (a – b) (a2+ab+b2)
(3a + b)³ – (2a + b)³ = (3a + b – 2a – b) [(3a+b)² + (3a + b)(2a + b) +(2a + b)²]
(3a + b)³ – (2a + b)³ = (3a + b – 2a – b) [(9a² + b² + 6ab) + (6a² +3ab +2ab +b²) +(4a² + b2 +4ab)]
(3a + b)³ – (2a + b)³ = (3a + b – 2a – b) (9a² +6a² +4a² +b² +b² +b² +6ab +3ab +2ab +4ab)
(3a + b)³ – (2a + b)³ = (3a + b – 2a – b) (19a² + 3b² + 15ab)
Hence, one factor of the given equation will be (3a + b – 2a – b) and the other factor will be (19a2 + 3b2 + 15ab).
Question 3: Find out the 5³2³?
Solution: As we know that the given equation is in the form of the a³b³ formula,
a³ – b³= (a – b) (a2+ab+b2)
Here a is indicated by 5 while b is indicated by 2
So, 5³2³ = (52) (25 + 10 + 4)
5³2³ = 3 (39) = 117 will be the answer.
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