Specific Term equals -1 for all variables. How to simplify?

Question

I am evaluating the expression $$\frac{(x^2-1)}{(1-x^2)}$$. When I plot this I see it is -1 for all x. This makes sense when I manually calculate it; however, given the expression is always equal to -1, is there not some way to mathematically cancel terms to show that it is equal to -1? I have tried simplifying it using $$\frac{(x-1)(x+1)}{(1-x)(1+x)}$$, but still cannot get terms to cancel out to show that it is equal to -1.

Answer 1

For algebraic manipulations, two common tricks is to add $0$ or multiply by $1$ somewhere in the expression. In either case, the final value of the expression doesn't change. But this allows you to rearrange the form.

Let's multiply by $1$: $$\frac{(x^2-1)}{(1-x^2)} = \frac{(x^2-1)}{(1-x^2)} \times 1$$

Now we know $\frac{-1}{-1} = 1$. So substituting for $1$, $$\frac{(x^2-1)}{(1-x^2)} \times 1 = \frac{(x^2-1)}{(1-x^2)} \times \frac{-1}{-1}$$

Now let's absorb the bottom $-1$ into the denominator of the first fraction $$\frac{(x^2-1)}{(1-x^2)} \times \frac{-1}{-1} = \frac{(x^2-1)}{(1-x^2) \times -1} \times (-1) = \frac{(x^2-1)}{(x^2 - 1)} \times (-1)$$

Now we see the numerator and denominator of the first fraction are same. So they cancel, as long as they are not $0$. $x^2-1 = 0 \rightarrow x = \pm 1$. So assuming $x\neq \pm 1$, $$\frac{(x^2-1)}{(x^2 - 1)} \times (-1) = 1 \times (-1) = -1$$

With practice, you will be able to quickly perform this cancellation in your head without having to go through all these pedantic steps, like in @MichalAdamaszek's comment