Review question

# If $x - 1/x = u$, what's $x^3 - 1/x^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9928

## Solution

If $x - \dfrac{1}{x} = u$, express $x^3 - \dfrac{1}{x^3}$ and $x^5 - \dfrac{1}{x^5}$ in terms of $u$.

We can begin by applying the binomial theorem to see that \begin{align*} \left( x - \frac{1}{x} \right)^3 &= x^3 - 3x + \frac{3}{x} - \frac{1}{x^3} \\ &= x^3 - \frac{1}{x^3} - 3\left( x - \frac{1}{x} \right) \end{align*} which implies that $\begin{equation*} x^3 - \frac{1}{x^3} = \left( x - \frac{1}{x} \right)^3 + 3 \left( x - \frac{1}{x} \right) = u^3 + 3u. \end{equation*}$ We can do a similar thing again: \begin{align*} \left( x - \frac{1}{x} \right)^5 &= x^5 - 5x^3 + 10x - \frac{10}{x} + \frac{5}{x^3} - \frac{1}{x^5} \\ &= x^5 - \frac{1}{x^5} - 5\left( x^3 - \frac{1}{x^3} \right) + 10 \left( x - \frac{1}{x} \right) \end{align*} This implies that \begin{align*} x^5 - \frac{1}{x^5} &= \left( x - \frac{1}{x} \right)^5 + 5 \left( x^3 - \frac{1}{x^3} \right) - 10 \left( x - \frac{1}{x} \right) \\ &= u^5 + 5(u^3 + 3u) - 10u \\ &= u^5 + 5u^3 + 5u. \end{align*}

If we extend this idea, we can write $x^7-\dfrac{1}{x^7}$, $x^9-\dfrac{1}{x^9}$ and so on as polynomials in $u$. Are there any patterns that you can find to the polynomials which result?