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?

Let $c = \cos \theta$.  You have $x^3 = c x - 1/2$, so $x^6 = (c x - 1/2)^2$ while $1/x = 2 c - 2 x^2$.  Then I get
$$\eqalign{x^6 + \dfrac{1}{2x^3} &= - 3 c^2 x^2 - 3 c x + 4 c^3 - \frac{3}{4}\cr & =
   -\dfrac{3}{2} (1 + \cos(2\theta)) x^2 - 3 \cos(\theta) x + \cos(3\theta) + 3 \cos(\theta) - \dfrac{3}{4}}$$
