Find Roots of $x^3-7$ Relative to Roots of Unity and $\sqrt{7}$

Question

find roots of $$x^3-7$$ relative to roots of unity and $\sqrt[3]7$

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one root is $\sqrt[3]7$, remaining roots might be something like $$\sqrt[3]7,(\sqrt[3]7)^2w,(\sqrt[3]7)^3w $$

where $w=e^{i 2 \pi/3 }$ [and wolram alpha is sayin $x_1=-\sqrt[3]{7},x_2 = \sqrt[3]{7},x_3=(-1)^{2/3} \sqrt[3]{7}$ ]

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Appreciate confirmation, whether this is wrong. or more elaboration. Trying to work out a problem in abstract algebra but my complex analysis is a bit rusted. Thanks.

Answer 1

It should be $\sqrt[3]{7},\sqrt[3]{7}\omega,\sqrt[3]{7}\omega^{2}$, where $\omega$ is the third root of unity, which is $\omega=\exp(2\pi i/3)$.

Answer 2

HINT: $$\begin{align}x^3-a^3&=(x-a)(x-a\omega)(x-a\omega^2)\end{align}$$

where $\omega$ is the cube root of unity.

Answer 3

The roots of unity are $1, \omega, \omega^2$ where $\omega = e^{i\frac 23 \pi}$

So $(1)^3, (\omega)^3, (\omega^2)^3$ all equal $1$.

So $(\sqrt[3]{7})^3,(\omega*\sqrt[3]{7})^3,(\omega^2*\sqrt[3]{7})^3$ will all equal $7$.

So $x = \sqrt[3]{7},\omega*\sqrt[3]{7},\omega^2*\sqrt[3]{7}$ are the three roots of $x^3 - 7 =0$.