<s>[INST] Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that 
$x^5 + \frac{1}{x^5} = 1$ as well.
 [/INST] Notice that: $$x + \frac{1}{x} = 1 \;\implies\; x^2 - x + 1 = 0 \;\implies\; x^3 + 1 = (x+1)(x^2-x+1) = 0 \;\implies\; x^3=-1$$
Then $x^6=1$ so: $$x^5 + \frac{1}{x^5} = \frac{x^6}{x} + \frac{x}{x^6} = \frac{1}{x} + x = 1$$
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