<s>[INST] 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.
 [/INST] As $x\ne0,$ on simplification we find $$x^2+1=\sqrt3x$$
Squaring we get $$x^4+2x^2+1=3x^2\iff x^4-x^2+1=0$$
Now
$$  x^6+1=(x^2+1)(x^4-x^2+1)$$
</s>