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I want to find the roots of the following equation: $z^6-3z^3+2$

Let's set $z^3=w$

We now have the polynomial: $w^2-3w+2=0$

$\Delta = 1 \implies w_1=1, w_2=2$

$z^3=w$ \implies:

$z_1= 1, z_2= $ $\sqrt[3]{2}$

I don't know how to find the other roots

Stefan4024
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aribaldi
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2 Answers2

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In $\mathbb{C}$, you know that $z^3=1$ has three solutions, as does $z^3=2$. This gives you the six solutions you are looking for.

Sloan
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You work is correct but not complete. For $z^3=1$ you have: $$ z^3-1=0 \quad \iff \quad (z-1)(z^2+z+1)=0 $$ so you have the solution $z=1$ but also $z=\frac{1}{2}(-1\pm i\sqrt{3})$ that are the roots of the second degree factor. And do the same for $z^3=2$.

Emilio Novati
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  • Once you know the roots of $z^3-1$, you can also simply multiply them by $\sqrt[3]{2}$. – Arnaud D. Sep 21 '16 at 15:58
  • How did you factorise into $(z-1)(z^2+z+1)=0$? how did you find the z^2+z+1 ? In this case is there an easier method than polynomial long division – aribaldi Sep 21 '16 at 16:02