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Find all the roots of the complex polynomial $p(z)=z^6+z^3-6$

I cant find any roots however and Im not sure how to simplify if I have a root to start with.

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    $\begingroup$ Not sure if it helps, but this is a quadratic in $z^3$. $\endgroup$ Sep 30, 2021 at 2:53
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    $\begingroup$ It will only have three complex conjugate pairs if there are no real roots. But $p(0)=-6,$ $p(2)>0, p(-2)>0.$ $\endgroup$ Sep 30, 2021 at 2:56
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    $\begingroup$ Or, write it as $\left(z^6-4\right)+\left(z^3-2\right)=\dots$ $\endgroup$
    – dxiv
    Sep 30, 2021 at 3:20

1 Answer 1

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You can factorise $P(z)=0$ as $(z^3-2)(z^3+3)=0$ then you need to find all three roots of $z^3=2$ and $z^3=-3$. Then you can write the rest of the roots just in cube root form like $z=(-3)^{\frac13}$ or $(-1)^{\frac23}2^{\frac13}$.

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