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?
