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 a³-b³ 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.
a3-b3 Formula
The formula of a3-b3 in algebra is a³-b³= (a-b) (a²+ab+b²). It is also known as a cube minus b cube formula. You can use the a3-b3 formula in maths to solve different algebraic problems.
a³-b³ = (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) |
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a3-b3 Formula Proof
There are various algebraic formulas in mathematics to be used for solving the equation. Here the complete proof derivation of the a3-b3 formula is given below.
Since we know the formula of (a-b)³= a³-b³-3ab(a-b)
a³-b³= (a-b)³+ 3ab(a-b)
a³-b³= (a-b) [(a-b)² + 3ab]
After the expansion of the equation on Right Hand Side (RHS),
a3-b³= (a – b) (a²+b²-2ab+3ab)
a3-b³= (a – b) (a²+b²+ab)
a3-b³= (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)
(a-b) (a²+ ab+ b²)= (a – b)(a² + ab + b²) = a3 + a2b + ab2 – a2b – ab2 – b3
Taking out the common factor (a-b),
(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,
(a-b) (a2+ ab+ b2)= a3-b3 (RHS) which is equal to Left Hand Side (LHS). Hence this formula is proved.
In this way, you should remember the a³-b³ formula to start your calculation in the examination.
a³-b³ = (a – b) (a²+ab+b²)
a³-b³ Formula Examples with Solutions
By memorizing this formula, you can easily calculate the mathematical problems.
Question 1: Factorize 125a³-27b³?
Solution: As we know that the expression of 125a³-27b³ can be written as (5a)³-(3b)³
By employing the a³-b³ formula,
a³-b³ = (a – b) (a²+ab+b²)
Now, we will get the factors of the given equation as,
(5a)³-(3b)³ = (5a-3b) (25a² + 15ab + 9b²)
Here 5a indicates a and 3b indicates b of the well-proven a³-b³ formula.
Question 2: Factorize (3a + b)³ – (2a + b)³?
Solution: As we know that the given equation is in the form of the a3-b3 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 a³-b³,
a3-b3= (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 5³-2³?
Solution: As we know that the given equation is in the form of the a³-b³ formula,
a³ – b³= (a – b) (a2+ab+b2)
Here a is indicated by 5 while b is indicated by 2
So, 5³-2³ = (5-2) (25 + 10 + 4)
5³-2³ = 3 (39) = 117 will be the answer.
