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

Jasonbradley

3 min read

·

Jun 16, 2024

30

Share

Factoring the Expression (a-b)³ + (b-c)³ + (c-a)³

This article will demonstrate how to factor the expression (a-b)³ + (b-c)³ + (c-a)³.

Understanding the Problem

We are given a cubic expression with three terms, each involving the difference of two variables raised to the power of three. Our goal is to rewrite this expression as a product of simpler expressions, which is called factoring.

Key Identity

To factor the given expression, we'll utilize the following algebraic identity:

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - xz - yz)

This identity holds true for any real numbers x, y, and z.

Applying the Identity

1. Relating the Expression to the Identity: Notice that the given expression resembles the left-hand side of the identity. We can make it match perfectly by introducing a term -3(a-b)(b-c)(c-a) and then subtracting it to maintain the equivalence:

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

2. Applying the Identity: Now, we can directly apply the identity with x = (a-b), y = (b-c), and z = (c-a):

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

3. Simplifying: The first term simplifies to 0. Expanding the second term and combining like terms, we get:

   3(a-b)(b-c)(c-a) = 3(a-b)(b-c)(c-a)

Final Result

Therefore, the factored form of the expression (a-b)³ + (b-c)³ + (c-a)³ is:

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

Conclusion

We successfully factored the expression (a-b)³ + (b-c)³ + (c-a)³ using a key algebraic identity. The resulting factored form is much simpler and easier to work with in various mathematical contexts.