(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+bc+ca-x^2

3 min read Jun 17, 2024

Factoring the Equation: (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+bc+ca-x^2

This equation presents an interesting challenge in algebraic manipulation, prompting us to explore the process of factoring and simplification. Let's break down the steps involved in proving this identity.

Expanding the Equation

The first step is to expand the left-hand side of the equation. We can do this by using the distributive property (or FOIL method) for each of the products:

(x-a)(x-b) = x² - ax - bx + ab
(x-b)(x-c) = x² - bx - cx + bc
(x-c)(x-a) = x² - cx - ax + ca

Now, let's combine these expanded terms:

x² - ax - bx + ab + x² - bx - cx + bc + x² - cx - ax + ca = ab + bc + ca - x²

Simplifying and Rearranging

Next, we combine like terms on the left-hand side:

3x² - 2ax - 2bx - 2cx + ab + bc + ca = ab + bc + ca - x²

Now, let's move the -x² term from the right-hand side to the left-hand side and simplify:

3x² - 2ax - 2bx - 2cx + ab + bc + ca + x² = ab + bc + ca

4x² - 2ax - 2bx - 2cx + ab + bc + ca = ab + bc + ca

Factoring by Grouping

We can now factor the left-hand side by grouping terms with common factors:

Group 1: 4x² - 2ax - 2bx - 2cx
Group 2: ab + bc + ca

Factoring out common factors in each group:

2x(2x - a - b - c) + (ab + bc + ca) = ab + bc + ca

Notice that the terms in the parentheses of the first group are the same as the terms in the second group. We can now factor out this common expression:

(2x - a - b - c)(2x + a + b + c) = 0

Conclusion

By expanding, simplifying, and factoring, we've successfully proven that the original equation is true. The equation is essentially a factored form of the expression ab + bc + ca - x², where the factors are (2x - a - b - c) and (2x + a + b + c). This demonstrates the power of algebraic manipulation in simplifying complex expressions and revealing underlying relationships.