<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] $x+\frac 1x=1\\
(x+\frac 1x)^2=1\\
x^2 + 2 + (\frac 1x)^2 = 1\\
x^2 + (\frac 1x)^2 = -1\\
x^4 + (\frac 1x)^4 = -1\\$
$x^5+(\frac 1x)^5 = (x+\frac 1x)(x^4 - x^3(\frac 1x) + x^2(\frac 1x)^2 - x(\frac 1x)^3 + (\frac 1x)^4)\\
(x+\frac 1x)(x^4 + (\frac 1x)^4 - x^2 - (\frac 1x)^2 + 1  )\\
 ( 1 )((-1) - (-1) + 1) = 1$
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