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. 

If $x^3 + \frac{1}{x^3} = 18$ then squaring gives $x^6 + 2 + x^{-6} = 18^2$ so $x^6 + x^{-6} = 322$.  Squaring again gives $x^{12} + x^{-12} = 322^2 - 2 = 103682$ 
Then $(x + x^{-1})(x^{11} + x^{-11}) = (x^{12} + x^{-12}) + (x^{10} + x^{-10})$. 
Now to me it seems clear that since we have 
$$
(x+ x^{-1})^3 = (x^3 + x^{-3}) + 3(x + x^{-1}),
$$ 
if we perform the substitution $t = x+ x^{-1}$ as suggested by @Remember_Me, then we get $t^3 - 3t - 18 = 0$. Now, doing the rational root test yields the root $t=3$, and then we can solve for the other two roots, but all choices will give the same result for $x^{11} + x^{-11}$, so we may as well take $x+ x^{-1} = 3$.
Then (squaring each time) we have $x^2 + x^{-2} = 7$, $x^{4} + x^{-4} = 47$ and $x^{8} + x^{-8} = 2207$ and so $(x^8 + x^{-8})(x^2 + x^{-2}) = (x^{10} + x^{-10}) + (x^6 + x^{-6}).$ Thus 
$$
x^{10} + x^{-10} = 2207 \times 7 - 322 = 15127,
$$
and so finally we get (from the above equation)
$$
3(x^{11} + x^{-11}) = 103682 + 15127 = 118809,
$$
so $x^11 + x^{-11} = 39603$, and we avoided...most...of the messy calculation!  
