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.
 A: $$t+\frac1t=18\sqrt3\iff t^2-(2\cdot9\sqrt3)t+1=0\iff t_{1,2}=\frac{9\sqrt3\pm\sqrt{(9\sqrt3)^2-1\cdot1}}1=$$
$$=9\sqrt3\pm\sqrt{81\cdot3-1}\quad=\quad9\sqrt3\pm\sqrt{243-1}\quad=9\sqrt3\pm\sqrt{242}\quad=\quad9\sqrt3\pm\sqrt{2\cdot121}=$$
$$=9\sqrt3\pm\sqrt{2\cdot11^2}\quad=\quad9\sqrt3\pm11\sqrt2\quad\iff\quad x_{1,2}^3=9\sqrt3\pm11\sqrt2=(a\sqrt3+b\sqrt2)^3=$$
$$=(a\sqrt3)^3+(b\sqrt2)^3+3(a\sqrt3)^2b\sqrt2+3a\sqrt3(b\sqrt2)^2\ =\ 3a^3\sqrt3+2b^3\sqrt2+9a^2b\sqrt2+6ab^2\sqrt3$$
$$\iff3a^3+6ab^2=9=3+6\quad,\quad2b^3+9a^2b=\pm11=\pm2\pm9\iff a=1,\quad b=\pm1.$$
A: 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
$$2\sqrt3=x+\frac{1}{x}\iff x^2-2\sqrt{3}x+1=0\iff (x-\sqrt3)^2-2=0 \iff x=\sqrt3\pm\sqrt2$$
A: $x^3+x^{-3}=18\sqrt{3}\Rightarrow x^6+1=18\sqrt{3}x^3\Rightarrow x^6-18\sqrt{3}x^3+1=0$
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.
A: $x^3+\frac{1}{x^3}=18\sqrt{3}\Rightarrow x^6-18\sqrt{3}x^3+1=0 $
we asssume $y=x^3$
$y^2-18\sqrt{3}y+1=0\Rightarrow y=\frac{18\sqrt{3}\pm\sqrt{968}}{2}=\frac{18\sqrt{3}\pm22\sqrt{2}}{2}=9\sqrt{3}\pm11\sqrt{2}$
$x^3-(9\sqrt{3}\pm11\sqrt{2})=0$ 
let $x=a\sqrt{3}+b\sqrt{2}$ now, 
$x^3=3\sqrt{3}a^3+2\sqrt{2}b^3+3.3a^2b\sqrt{2}+3.a\sqrt{3}.2b^2=(3a^3+6ab^2)\sqrt{3}+(2b^3+9a^2b)\sqrt{2}$
Now, $3a^3+6ab^2=9$ and $2b^3+9a^2b=11$ 
i.e., $a^3+2ab^2=3$ and $2b^3+9a^2b=11$.
Now, 
for $a^3+2ab^2=3$ one possible case would be $a^3=1,ab^2=1$ (I am not saying this is "the" one)
i.e., $a=1 ,b=\pm 1$
Now, 
for $2b^3+9a^2b=11$ one possible case would be $b^3=1,a^2b=1$ (I am not saying this is "the" one)
i.e., $b=1, a=\pm 1$
but then we need both $a^3+2ab^2=3$ and $2b^3+9a^2b=11$ to satisfy at once.
So, only possibility would be $a=1,b=1$ which would imply $x=\sqrt{3}+\sqrt{2}$
Note : I took only $+11$ in $9\sqrt{3}\pm11\sqrt{2}$ , I would leave it to you to do the same for $-11$ case.
Good Luck!
