Review question

# Can we simplify these algebraic fractions? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5364

## Solution

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*}