Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$ Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$
Hints are appreciated. Thanks in advance.

Another way:
$$x^{18} + x^{24} = x^{21} (x^3 + \frac{1}{x^3})\\ = x^{21} ((x+\frac{1}{x})^3 - 3(x+\frac{1}{x}))\\ = x^{21}(3\sqrt{3} - 3\sqrt{3}) = 0. $$
