Simplifying 3rd root of (24 * sqrt(3))

Question

I have problems following a solution towards simplifying a given polynomial.

$$Polynomial: p(x)=x^5+{\sqrt 3}x^4+24{\sqrt 3}x^2+72x$$

the zeros of this function (Polynomial roots? English isn't my native language, so I don't know how to express the point(s) at which the function meets the X-axis) are: $$x_0 = -{\sqrt 3} \\ x_1 = 0 \\$$ and the complex ones, which are calculated with what "remains" after polynomial division, drawing the 3rd root, etc.: $$x=\sqrt[\Large 3]{-24\sqrt 3}$$

The first problem comes now. The next step, without explanation, simplifies the above to the following: $$x=\sqrt[\Large 3]{8*\sqrt 3^3e^{\Large_{i\pi}}}$$ How is this done or rather what's the logic behind it? Especially the 8 that somehow was transformed from the 24.

Answer 1

$24 = 3*8$ so $24 \sqrt {3} = 8*3*\sqrt{3} = 8\sqrt{3}^3$

$-1 = e^{\pi i}$ so $-24\sqrt{3} = 8*(\sqrt 3)^3 e^{\pi i}=2^3(\sqrt 3)^3e^{\pi i}$

And so $\sqrt[3]{-24\sqrt 3}=\sqrt[3]{2^3\sqrt{3}^3e^{\pi i}} = 2\sqrt 3 e^{\frac {(2k + 1)}3\pi i}$

Answer 2

The $8$ wasn't "transformed from" the $24$. It was factored: $24=8\cdot3=8\cdot\sqrt 3^2$, so $24\sqrt3=8\cdot\sqrt3^3$. The $-1$ became rewritten as $e^{\pi i}$.