<s>[INST] If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$. I have the following question:

If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.

Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx-1)^{1.5}$ for $x^3$ in the second equation. Ideas?
 [/INST] I don't see any way to continue from what you've done, since you're dealing with fractional powers of a polynomial, which are generally hard to work with. Instead, since $x \neq 0$, we can divide both sides by $x$ below to get
$$x^2 + 1 = kx \;\; \to \; \; x+\frac{1}{x}=k$$
Cubing both sides then gives
$$\begin{equation}\begin{aligned}
x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} & = k^3 \\
x^3 + 3\left(x + \frac{1}{x}\right) + \frac{1}{x^3} & = k^3 \\
x^3 + 3k + \frac{1}{x^3} & = k^3 \\
x^3 + \frac{1}{x^3} & = k^3 - 3k
\end{aligned}\end{equation}$$
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