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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$\begingroup$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 $a+ib\neq 1$.
$\endgroup$ $\begingroup$If $z^3=w^3$, then $$\left(\frac{z}{w}\right)^3=1\\\left(\frac{z}{w}\right)^3=e^{2\pi ki}\\z=we^{\frac{2\pi i k}{3}}$$ That is, $z,w$ differ by a unit. In particular, there are three possibilities:
- $a+bi=(x+iy)\cdot 1$
- $a+bi=(x+iy)\cdot e^{\frac{2\pi i}{3}}$
- $a+bi=(x+iy)\cdot e^{\frac{4\pi i}{3}}$
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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