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.
Expanding the Expressions
Let's start by expanding each term in the given expression:
$(x-a)(x-b) = x^2 - bx - ax + ab$
$(x-b)(x-c) = x^2 - cx - bx + bc$
$(x-c)(x-a) = x^2 - ax - cx + ac$
Adding the Expanded Terms
Now, let's add the expanded terms together:
$x^2 - bx - ax + ab + x^2 - cx - bx + bc + x^2 - ax - cx + ac$
Combining Like Terms
Next, we'll combine like terms:
$3x^2 - 3ax - 3bx + ab + bc + ac$
Factoring Out the Common Terms
Notice that we can factor out a common term of $(x - a)$
:
$(x - a)(3x - 3b + a) + bc + ac$
Simplifying the Expression
Now, we can simplify the expression by combining the remaining terms:
$(x - a)(3x - 3b + a) + b(c - a)$
The Final Answer
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`$.