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Does $(a + bi)^3 = (x+yi)^3 \implies a + bi = x + yi$, where $a,b,x,y \in \mathbb{R}?$

Please provide a proof/counterexample. I've tried brute force and ended up with:

$$(a - x)(a^2 + ax + x^2) + 3(-ab^2 + xy^2) = 0$$

$$(y - b)(y^2 + by + b^2) + 3(a^2b - x^2y) = 0$$

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    $\begingroup$ what are $c$ and $d$? $\endgroup$ – asd Feb 2 '19 at 11:12
  • $\begingroup$ @asd Sorry, I meant $x,y$ $\endgroup$ – user168651 Feb 2 '19 at 11:13
  • $\begingroup$ you mean $(a+ib)^3 = (x+yi)^3 \Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false... $\endgroup$ – dfnu Feb 2 '19 at 11:16
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Take $a+ib=e^{\frac{2\pi i}{3}}$ and $x+iy=1$ We have $$ (a+ib)^3=(e^{\frac{2\pi i}{3}})^3=e^{2\pi i}=1=1^3 $$ but $a+ib\neq 1$.

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If $z^3=w^3$, then $$z^3=(we^{2\pi ik})^3\\z=we^{\frac{2\pi ik}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:

  • $a+bi=(x+iy)\cdot 1$
  • $a+bi=(x+iy)\cdot e^{\frac{2\pi i}{3}}$
  • $a+bi=(x+iy)\cdot e^{\frac{4\pi i}{3}}$
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No. There are $3$ roots of $z^3=z_0$ for any $0\not=z_0\in\Bbb C$.

If $\gamma$ is one, then $\sigma\gamma,\sigma ^2\gamma $ are the other two, where $\sigma =e^{\frac{2\pi i}3}$.

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