Simplify the expression \[
\frac{1}{x^2 - x} - \frac{1}{x^3 - x}.
\]
Each denominator has a factor
\(x\), so we can write
\[
\frac{1}{x^2 - x} - \frac{1}{x^3 - x} = \frac{1}{x(x-1)} - \frac{1}{x(x^2-1)}.
\] Next, we can write
\(x^2 - 1 = (x-1)(x+1)\), and so
\[\begin{align*}
\frac{1}{x(x-1)} - \frac{1}{x(x^2-1)} &= \frac{1}{x(x-1)} - \frac{1}{x(x-1)(x+1)} \\
&= \frac{x+1}{x(x-1)(x+1)} - \frac{1}{x(x-1)(x+1)} \\
&= \frac{x}{x(x-1)(x+1)} \\
&= \frac{1}{(x-1)(x+1)}.
\end{align*}\]
We know that \(x\) is not \(0, 1\) or \(-1\), since in these cases, the question makes no sense.
Find for what values of \(x\) this expression is equal to \(1\).
By using the simplified form we found above, this is equivalent to asking us to find the values of
\(x\) for which
\[
\frac{1}{(x-1)(x+1)} = 1.
\] By rearranging, we have that
\[\begin{align*}
1 = (x-1)(x+1) &\iff 1 = x^2 - 1 \\
&\iff x^2 - 2 = 0 \\
&\iff x = \pm\sqrt{2}.
\end{align*}\]
