# Does $(a + bi)^3 = (x+yi)^3 \implies a + bi = x + yi$?

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:

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

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

• what are $c$ and $d$? – asd Feb 2 '19 at 11:12
• @asd Sorry, I meant $x,y$ – user168651 Feb 2 '19 at 11:13
• you mean $(a+ib)^3 = (x+yi)^3 \Rightarrow a+ib = x+iy$? If this is the case, the assertion is clearly false... – dfnu Feb 2 '19 at 11:16

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 $$a+ib\neq 1$$.
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:
• $$a+bi=(x+iy)\cdot 1$$
• $$a+bi=(x+iy)\cdot e^{\frac{2\pi i}{3}}$$
• $$a+bi=(x+iy)\cdot e^{\frac{4\pi i}{3}}$$
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 $$\sigma =e^{\frac{2\pi i}3}$$.