Problem with $x^{6} - 2 = 0$ compute roots in $\mathbb{C}$ 
I have problem with simple equation $x^{6} - 2 = $ compute roots in
  $\mathbb{C}$

I will try compute roots of $x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=(x-2^{1/6})(x^{2}+2^{1/6}x+2^{1/3})(x^{3}+\sqrt{2})$, but this not looks good. 
Maybe is better solution. 
Any suggestions? 

As commented above, you used $a^3-b^3=(a-b)(a^2+ab+b^2)$. You can also use $a^3+b^3=(a+b)(a^2-ab+b^2)$ and easily solve the quadratic equations:
$$x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=\color{red}{(x-2^{1/6})}\color{green}{(x^{2}+2^{1/6}x+2^{1/3})}\color{blue}{(x+2^{1/6})}\color{purple}{(x^{2}-2^{1/6}x+2^{1/3})} \Rightarrow \\
\color{red}{x_1=2^{1/6}},\color{blue}{x_2=-2^{1/6}},\\
\color{green}{x^2+2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{-2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{-2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2},\\
\color{purple}{x^2-2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2}.
$$
