(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 Are Always

2 min read Jul 03, 2024

Factoring and Proving a Mathematical Identity

In this article, we will prove the mathematical identity:

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

This identity is often referred to as a "always true" or "identically equal to zero" in mathematical terms.

Let's start by expanding each term in the given expression:

$(x-a)(x-b) = x^2 - bx - ax + ab$

$(x-b)(x-c) = x^2 - cx - bx + bc$

$(x-c)(x-a) = x^2 - ax - cx + ac$

Now, let's add the expanded terms together:

$x^2 - bx - ax + ab + x^2 - cx - bx + bc + x^2 - ax - cx + ac$

Next, we'll combine like terms:

$3x^2 - 3ax - 3bx + ab + bc + ac$

Notice that we can factor out a common term of $(x - a)$ :

$(x - a)(3x - 3b + a) + bc + ac$

Now, we can simplify the expression by combining the remaining terms:

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

Finally, we can see that the original expression is indeed identically equal to zero:

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

This mathematical identity is always true for any values of $x``, $a$, $b$, and $c`$.