<s>[INST] What are the linear factors of $x^3 - 2$? The polynomial is defined as $$x^3 - 2 \in \mathbb{C}[x]$$
How do you figure out the linear factors of the term? If I evaluate the function in polar coordinates I get some weird terms that can't possibly be correct. Wolfram says that I have to multiply the real solution by the third roots of $i$ (as that's the unity vector), but if I write those solutions as linear factors and multiply the term out I get something different from my polynomial. 
Can someone please help me? 
Thank you! 
Edit: My attempt at solving with polar coordinates: $$x^3 = 2 = 2+0i = 2(\cos(0) + i\sin(0))  | \sqrt[3]{}$$ $$x = \sqrt[3]{2}(\cos(L*\frac{2\pi}{3}) + i\sin(L*\frac{2\pi}{3}) | L \in \{0, 1, 2\}$$ $$x_1 = \sqrt[3]{2}$$ $$x_2 = \sqrt[3]{2}(-\frac{1}{2} + i\frac{\sqrt{3}}{2})$$ $$x_3 = \sqrt[3]{2}(-\frac{1}{2} - i\frac{\sqrt{3}}{2})$$
 [/INST] See that any root that you could have of $x^3-2$ is of the style $\sqrt[3]{2}\ e^{2it\pi/3}$ with $t=0,1,2$, so it factors into
$$(x-\sqrt[3]{2})(x-\sqrt[3]{2}\ e^{2i\pi/3})(x-\sqrt[3]{2}\ e^{4i\pi/3})$$
If you want to know what is $e^{2ti\pi/3}$ use the polar coordinates:
$$e^{2it\pi/3}=\cos\left(\frac{2\pi t}{3}\right)+i\sin\left(\frac{2\pi t}{3}\right)$$
And all the roots are:
$$\sqrt[3]{2}\ e^{2i\pi/3}=\sqrt[3]{2}\left( \cos\left(\frac{2\pi }{3}\right)+i\sin\left(\frac{2\pi }{3}\right)\right)=\sqrt[3]{2}\left(-\frac{1}{2}+-\frac{i\sqrt{3}}{2}\right)$$
$$\sqrt[3]{2}\ e^{4i\pi/3}=\sqrt[3]{2} \left(\cos\left(\frac{4\pi }{3}\right)+i\sin\left(\frac{4\pi }{3}\right)\right)=\sqrt[3]{2}\left(-\frac{1}{2}--\frac{i\sqrt{3}}{2}\right)$$
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