(a-b)3+(b-c)3+(c-a)3

3 min read Jun 16, 2024
(a-b)3+(b-c)3+(c-a)3

Exploring the Identity: (a-b)³ + (b-c)³ + (c-a)³

This article delves into the intriguing algebraic identity: (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b)(b-c)(c-a). We'll explore its proof, applications, and why it holds significance in mathematics.

Understanding the Identity

The identity states that the sum of the cubes of the differences between three variables, a, b, and c, is equal to three times the product of those differences. This seemingly complex expression has a surprising and elegant result.

Proof

The proof involves expanding the cubes and simplifying the expression:

  1. Expand the cubes: (a-b)³ = a³ - 3a²b + 3ab² - b³ (b-c)³ = b³ - 3b²c + 3bc² - c³ (c-a)³ = c³ - 3c²a + 3ca² - a³

  2. Add the expanded expressions: (a-b)³ + (b-c)³ + (c-a)³ = (a³ - 3a²b + 3ab² - b³) + (b³ - 3b²c + 3bc² - c³) + (c³ - 3c²a + 3ca² - a³)

  3. Cancel out terms: Notice that the cubic terms (a³, b³, c³) cancel out.

  4. Factor out common terms: = -3a²b + 3ab² - 3b²c + 3bc² - 3c²a + 3ca² = 3(ab² - a²b - b²c + bc² - c²a + ca²)

  5. Further factorization: = 3(a-b)(b-c)(c-a)

Therefore, the identity (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b)(b-c)(c-a) is proven.

Applications

This identity is useful in various areas of mathematics, including:

  • Algebraic manipulation: Simplifying complex expressions involving differences of cubes.
  • Number theory: Studying properties of integers and their relationships.
  • Geometry: Solving problems related to volumes and surfaces.

Significance

The identity highlights the power of factorization and simplification in mathematics. It showcases how seemingly complex expressions can be reduced to simpler forms, revealing underlying relationships between variables.

Conclusion

The identity (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b)(b-c)(c-a) is a valuable tool for mathematicians and anyone working with algebraic expressions. Its elegance and practical applications solidify its significance in the world of mathematics.