<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] $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)\\=(x+\frac{1}{x})[(x+\frac{1}{x})^2-3]$$
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