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.

$$t+\frac1t=18\sqrt3\iff t^2-(2\cdot9\sqrt3)t+1=0\iff t_{1,2}=\frac{9\sqrt3\pm\sqrt{(9\sqrt3)^2-1\cdot1}}1=$$
$$=9\sqrt3\pm\sqrt{81\cdot3-1}\quad=\quad9\sqrt3\pm\sqrt{243-1}\quad=9\sqrt3\pm\sqrt{242}\quad=\quad9\sqrt3\pm\sqrt{2\cdot121}=$$
$$=9\sqrt3\pm\sqrt{2\cdot11^2}\quad=\quad9\sqrt3\pm11\sqrt2\quad\iff\quad x_{1,2}^3=9\sqrt3\pm11\sqrt2=(a\sqrt3+b\sqrt2)^3=$$
$$=(a\sqrt3)^3+(b\sqrt2)^3+3(a\sqrt3)^2b\sqrt2+3a\sqrt3(b\sqrt2)^2\ =\ 3a^3\sqrt3+2b^3\sqrt2+9a^2b\sqrt2+6ab^2\sqrt3$$
$$\iff3a^3+6ab^2=9=3+6\quad,\quad2b^3+9a^2b=\pm11=\pm2\pm9\iff a=1,\quad b=\pm1.$$
