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.

Set $a=x+\frac{1}{x}, b\sqrt3=a$. Then we want that
$$a^3-3a=18\sqrt3\iff b^3-b=6\iff(b-2)((b+1)^2+2)=0\iff b=2$$
So
$$2\sqrt3=x+\frac{1}{x}\iff x^2-2\sqrt{3}x+1=0\iff (x-\sqrt3)^2-2=0 \iff x=\sqrt3\pm\sqrt2$$
