a³ - b³ Formula: Definition, Proof, and Solved Examples

Algebra is a vital branch of mathematics that simplifies complex problems using variables such as x, y, and z. In this guide, we will break down the algebraic formula of a³-b³, including its step-by-step proof, derivation, and practical applications. Mastering this formula is essential for students and those preparing for competitive exams like NTSE, NDA, AFCAT, SSC, and Railway recruitment.

a3-b3 Formula

The algebraic formula for the difference of two cubes is a³-b³ = (a-b)(a²+ab+b²). Commonly referred to as the cube minus cube formula, this identity is a fundamental tool for solving a wide variety of algebraic expressions and polynomial factorizations.

a³ - b³ = (a - b) (a² + ab + b²)

Basics of Algebraic Equation

In mathematics, an algebraic equation is a statement asserting the equality of two expressions. These equations are constructed using a combination of variables, coefficients, and constants.

An algebraic equation provides a balanced relationship between unknown variables and constants. Because both sides of the equation maintain equal value, it is considered a balanced identity, often represented in the form P=0, where P is a polynomial.

a3 b3 Formula Chart

There are several related identities in algebra involving a³ and b³. Let's examine the comprehensive a³ ± b³ formula chart.

a3-b3 Formula Proof

Mathematics relies on various identities to simplify and solve equations efficiently. Below, you will find the complete step-by-step derivation for the a³-b³ formula.

Starting from the known cube expansion formula: (a-b)³ = a³ - b³ - 3ab(a-b)

Rearranging the terms: a³ - b³ = (a-b)³ + 3ab(a-b)

Factoring out (a-b): a³ - b³ = (a-b) [(a-b)² + 3ab]

Expanding the term within the square brackets on the Right Hand Side (RHS):

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

Combining like terms: a³ - b³ = (a - b) (a² + b² + ab)

Final factored form: a³ - b³ = (a - b) (a² + ab + b²)

Alternatively, we can verify the formula by expanding the factors on the right side:

Taking the Right Hand Side (RHS):

(a-b)(a² + ab + b²) = a(a² + ab + b²) – b(a² + ab + b²)

(a-b)(a² + ab + b²) = a³ + a²b + ab² – a²b – ab² – b³

Grouping and simplifying terms:

(a-b)(a² + ab + b²) = a³ + (a²b – a²b) + (ab² – ab²) – b³

Canceling the terms a²b and ab²:

(a-b)(a² + ab + b²) = a³ - b³ (RHS), which equals the Left Hand Side (LHS). The identity is now proven.

Memorizing the a³-b³ formula is essential for speed and accuracy during your examinations.

a³ - b³ = (a - b) (a² + ab + b²)

a³-b³ Formula Examples with Solutions

Internalizing this identity allows you to solve complex algebraic problems quickly.

Question 1: Factorize 125a³ - 27b³

Solution: Note that 125a³ - 27b³ can be rewritten as (5a)³ - (3b)³.

Using the standard a³ - b³ identity:

a³ - b³ = (a - b) (a² + ab + b²)

Applying this to our values where a = 5a and b = 3b:

(5a)³ - (3b)³ = (5a - 3b) (25a² + 15ab + 9b²)

This result represents the factored form of the expression.

Question 2: Factorize (3a + b)³ - (2a + b)³

Solution: The expression follows the structure of the a³ - b³ formula.

Let a = (3a + b) and b = (2a + b).

Apply the algebraic identity:

a³ - b³ = (a - b) (a² + ab + b²)

Substituting the expressions:

(3a + b)³ - (2a + b)³ = [(3a + b) - (2a + b)] * [(3a+b)² + (3a+b)(2a+b) + (2a+b)²]

Expanding the terms: = (a) * [(9a² + b² + 6ab) + (6a² + 3ab + 2ab + b²) + (4a² + b² + 4ab)]

Combining like terms: = (a) * (19a² + 3b² + 15ab)

The factored form is a(19a² + 15ab + 3b²).

Thus, the expression is factored into the product of 'a' and the resulting quadratic expression.

Question 3: Calculate the value of 5³ - 2³

Solution: This matches the structure of the a³ - b³ formula.

Where a = 5 and b = 2.

Using the formula (a - b)(a² + ab + b²):

5³ - 2³ = (5 - 2)(5² + (5 * 2) + 2²)

5³ - 2³ = 3 * (25 + 10 + 4) = 3 * 39 = 117.