<s>[INST] Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me. 
 [/INST] Our old friend the golden ratio, or rather its enhancement by $1$, namely $a=1+\phi$, along with its reciprocal, is lurking behind this problem.
Let $a$ and $\bar a$ be respectively the larger and smaller root of $x^2-3x+1=0$. We have $a\bar a=1$ and $a+\bar a=3$. It's easy to show that $a^3+\bar a^3=18$. Also,$$a^{11}+\bar a^{11}=(a^5+\bar a^5)(a^6+\bar a^6)-(a+\bar a).$$After breaking $a^5+\bar a^5$ down to $(a^3+\bar a^3)[(a+\bar a)^2-2]-(a+\bar a)$, and $a^6+\bar a^6$ to $(a^3+\bar a^3)^2-2$, we get the answer$$a^{11}+\bar a^{11}=(18\times7-3)(18^2-2)-3=39,\!603.$$(Here we are ignoring the complex roots.)
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