If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$ 
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$

The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take  1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means?
NOTE: This problem is from a textbook of mine.

$x^3+x^{-3}=18\sqrt{3}\Rightarrow x^6+1=18\sqrt{3}x^3\Rightarrow x^6-18\sqrt{3}x^3+1=0$
From there if you look at it like a quadratic equation, you can find 2 solutions for $x^3$  So x would simply be the cube roots of that. and $x^3=a$ has only 1 real root for any real-valued a.
