Answer
$w_0=\sqrt[6]{2}\left[\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right]$
$w_1=\sqrt[6]{2}\left[\cos\left(\frac{11\pi}{12}\right)+i\sin\left(\frac{11\pi}{12}\right)\right]$
$w_2=\sqrt[6]{2}\left[\cos\left(\frac{19\pi}{12}\right)+i\sin\left(\frac{19\pi}{12}\right)\right]$
$w_3=\sqrt[6]{2}\left[\cos\left(\frac{5\pi}{12}\right)+i\sin\left(\frac{5\pi}{12}\right)\right]$
$w_4=\sqrt[6]{2}\left[\cos\left(\frac{13\pi}{12}\right)+i\sin\left(\frac{13\pi}{12}\right)\right]$ $w_5=\sqrt[6]{2}\left[\cos\left(\frac{7\pi}{4}\right)+i\sin\left(\frac{7\pi}{4}\right)\right]$
Work Step by Step
Consider the equation $z^6+2z^3+2=0$.
Observe that
$z^6+2z^3+2=0\Longrightarrow$
$\left(z^3\right)^2+2z^3+1+1=0\Longrightarrow$
$\left(z^3+1\right)^2+1=0\Longrightarrow$
$\left(z^3+1\right)^2=-1\Longrightarrow$
$z^3+1=-i$ or $z^3+1=i$.
Thus, $z^3=-1-i$ or $z^3=-1+i$. and therefore, $z=\sqrt[3]{-1-i}$ or $z=\sqrt[3]{-1+i}$.
Now, let's calculate the cubic roots above.
Observe that $$\left|-1-i\right|=\sqrt{\left(-1\right)^2+\left(-1\right)^2}=\sqrt{1+1}=\sqrt{2}$$ and $$\left|-1+i\right|=\sqrt{\left(-1\right)^2+1^2}=\sqrt{1+1}=\sqrt{2}.$$
Moreover, the polar angle corresponding to $-1-i$ is $\displaystyle\frac{5\pi}{4}$ and the polar angle corresponding to $-1+i$ is $\displaystyle\frac{3\pi}{4}$ which implies that $$-1-i=\sqrt{2}\left[\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)\right]=\sqrt{2}\exp\left(i\left(\frac{5\pi}{4}\right)\right)$$ and $$-1+i=\sqrt{2}\left[\cos\left(\frac{3\pi}{4}\right)+i\sin\left(\frac{3\pi}{4}\right)\right]=\sqrt{2}\exp\left(i\left(\frac{3\pi}{4}\right)\right).$$
Thus, $\sqrt[3]{-1-i}=\sqrt[3]{\sqrt{2}\exp\left(i\left(\frac{5\pi}{4}\right)\right)}=\sqrt[3]{\sqrt{2}}\exp\left[i\left(\displaystyle\frac{\left(\frac{5\pi}{4}\right)}{3}+k\displaystyle\frac{2\pi}{3}\right)\right]=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{5\pi}{12}+k\displaystyle\frac{2\pi}{3}\right)\right]=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{8k+5}{12}\right)\pi\right]$ where $0\leq k\leq 2$ and $\sqrt[3]{-1+i}=\sqrt[3]{\sqrt{2}\exp\left(i\left(\frac{3\pi}{4}\right)\right)}=\sqrt[3]{\sqrt{2}}\exp\left[i\left(\displaystyle\frac{\frac{3\pi}{4}}{3}+k\displaystyle\frac{2\pi}{3}\right)\right]=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{\pi}{4}+k\displaystyle\frac{2\pi}{3}\right)\right]=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{8k+3}{12}\right)\pi\right]$ where $0\leq k\leq 2$.
Hence, the $6$ solutions of the equation are: $$w_0=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{\pi}{4}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{\pi}{4}\right)+i\sin\left(\displaystyle\frac{\pi}{4}\right)\right],$$ $$w_1=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{11\pi}{12}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{11\pi}{12}\right)+i\sin\left(\displaystyle\frac{11\pi}{12}\right)\right],$$ $$w_2=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{19\pi}{12}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{19\pi}{12}\right)+i\sin\left(\displaystyle\frac{19\pi}{12}\right)\right],$$ $$w_3=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{5\pi}{12}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{5\pi}{12}\right)+i\sin\left(\displaystyle\frac{5\pi}{12}\right)\right],$$ $$w_4=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{13\pi}{12}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{13\pi}{12}\right)+i\sin\left(\displaystyle\frac{13\pi}{12}\right)\right],$$ $$w_5=\sqrt[6]{2}\exp\left[i\left(\displaystyle\frac{7\pi}{4}\right)\right]=\sqrt[6]{2}\left[\cos\left(\displaystyle\frac{7\pi}{4}\right)+i\sin\left(\displaystyle\frac{7\pi}{4}\right)\right].$$
