Sum Of Cubes Will Be Handy

$Let ~\color{#007fff}{x=\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{2}}}$ $\small{\color{#D61F06}{(Using (a+b)^3=a^3+b^3+3ab(a+b))}}$ $\Rightarrow x^3=2+\dfrac{1}{2}+3(\color{#007fff}{\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{2}}})$ $\Rightarrow x^3=2+\dfrac{1}{2}+3(\color{#007fff}{x})$ $\Rightarrow x^3-3x=\dfrac{5}{2}$ $\Rightarrow 2x^3-6x=\huge\boxed{\color{#0C6AC7}{5}}$

Let $a = \sqrt[3]2$ . Then $x = a + \dfrac 1a$ and:

$\begin{aligned} 2x^3 - 6x & = 2\left(a+\frac 1a\right)^3 - 6\left(a+\frac 1a\right) \\ & = 2\left(a^3+3a + \frac 3a + \frac 1{a^3} \right) - 6\left(a+\frac 1a\right) \\ & = 2\left(a^3 + \frac 1{a^3} \right) \\ & = 2\left(2 + \frac 12 \right) \\ & = \boxed{5} \end{aligned}$

$2x^3 - 6x$ $= 2x(x^2-3)$ $= (2^{{4}/{3}}+2^{{2}/{3}})(2^{{2}/{3}}+2^{-{2}/{3}}-1)$ $=2^2 + 2^{{2}/{3}} - 2^{{4}/{3}} + 2^{{4}/{3}} + 1 - 2^{{2}/{3}}$ $= \large \color{#20A900}{\boxed{5}}$