how do we get complex roots for $x^3 = 8$? I just asked myself. if you have:
$$x^3 -8 = 0$$
and trying to solve it, you do
$$x^3 = 8$$
well, the first real root will be $2$ because $2^3 = 8$
but it is only the first, what about second and third?
 A: HINT: use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
A: We know  $x = 2$ is one root, so $f (x) $ has a factor of $(x-2) $. Simplifying, we have, $$x^3 = 8$$ $$\implies x^3-8=0$$ $$\implies (x-2)(x^2+2x+4) = 0$$
Now you can solve the quadratic by the quadratic formula. Its roots are complex.
A: If $x\in\mathbb C$ is such that $x^3=8$, then $|x|^3=|x^3|=8$ and therefore $|x|=2$. So, write $x$ as $2(\cos\theta +i\sin\theta)$ ($\theta\in[0,2\pi)$). Then $x^3=8\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)$. When is this equal to $8$? That's when $\theta=0$, $\theta=\frac{2\pi}3$ or $\theta=\frac{4\pi}3$.
A: $$x^3-8=0$$
One of the root is $2$. The complex root comes in pair when the coefficient is real.
From Vieta's formula,
$$2+(a+ib)+(a-ib)=0 \implies 2a+2=0 \implies a=-1$$
$$2(a+ib)(a-ib)=8 \implies a^2+b^2=4$$
You just have to solve for $b$.
A: Let $x=u+iv$. Then cubing and equating real/imaginary parts,
$$u^3-3uv^2=8,\\3u^2v-v^3=0.$$
The second equation tells us that $v=0$ or $v=\pm\sqrt3u$. And by the first,


*

*$v=0\to x=2$,

*$v=\sqrt u\to -8u^3=8,x=-1-i\sqrt3$,

*$v=-\sqrt u\to -8u^3=8,x=-1+i\sqrt3$.
(Note that only the real $u,v$ solutions of these equations can be used and there is no circular argument.)
