<s>[INST] If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
 [/INST] $$2x^3+1=2x\cos(a)\implies x^3-x\cos(a)+{1\over 2}=0$$Let $x=y+{b\over y}\implies y={1\over 2}(x+\sqrt{x^2-4b})$ Using back substitution$$\left(y+{b\over y}\right)^3-\left(y+{b\over y}\right)\cos(a)+{1\over 2}=0$$$$\implies {1\over 2}y^3+y^6+b^3+y^4(3b-\cos(a))+y^2(3b^2-b\cos(a))=0$$
Now substitute $b={\cos(a)\over 3}$ and $z=y^3$ $$z^2+{z\over 2}+{\cos^3(a)\over 27}=0\implies z={1\over 36}\left(\sqrt{3}\sqrt{27-16\cos^3(a)}-9\right)$$
Now substitute back $z=y^3$ and it will give the $3$ values of $y$.
Once you get the values of $y$ substitute back the values of $x$ and you will get $3$ quadratics  of $x$ 
BINGO! 
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