(a-b)3+(b-c)3+(c-a)3 Is Equal To

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Jul 03, 2024

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(a-b)3 + (b-c)3 + (c-a)3 is equal to 0

In algebra, there is a fascinating identity that involves the sum of three terms, each of which is a cube of a difference of variables. Specifically, the identity states that:

(a-b)3 + (b-c)3 + (c-a)3 = 0

This identity may seem surprising at first, but it can be proven through a series of simple algebraic manipulations.

Proof

Let's start by expanding each of the terms in the sum:

(a-b)3 = a3 - 3a2b + 3ab2 - b3 (b-c)3 = b3 - 3b2c + 3bc2 - c3 (c-a)3 = c3 - 3c2a + 3ca2 - a3

Now, let's add up these three terms:

(a-b)3 + (b-c)3 + (c-a)3 = (a3 - 3a2b + 3ab2 - b3) + (b3 - 3b2c + 3bc2 - c3) + (c3 - 3c2a + 3ca2 - a3)

Combine like terms:

a3 - 3a2b + 3ab2 - b3 + b3 - 3b2c + 3bc2 - c3 + c3 - 3c2a + 3ca2 - a3

Notice that many of the terms cancel out, leaving us with:

-3a2b + 3ab2 - 3b2c + 3bc2 - 3c2a + 3ca2 = 0

This equation is true for all values of a, b, and c, which means that the original identity is indeed true:

(a-b)3 + (b-c)3 + (c-a)3 = 0

This identity has many applications in mathematics, particularly in algebra and number theory. It is a powerful tool for simplifying complex expressions and solving equations.