How to find the complex roots of $y^3-\frac{1}{3}y+\frac{25}{27}$ I've been trying to solve this for hours and all  found was the real solution by Cardano"s formula.
I vaguely remember that if $\alpha$ is a root of a complex number, the other roots are $\omega \alpha$ and $\omega ^2 \alpha$ where $\omega = - \frac{1}{2}+i \frac{\sqrt{3}}{2}$, $\omega ^2 = - \frac{1}{2}-i \frac{\sqrt{3}}{2}$.
The problem is, I don't know how to apply this to find my roots(if applicable at all).
Straightly put, how can I find other roots of a cubic given one root?(This time it happens to be around -0.75, for the real root). Given a single root, what is the relationship between others? Any formula to compute the remaining solutions at all?
 A: By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;
and replacing $a, b, c$ by $x, -u, -v$ respectively,
$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.
$u, v$ are known as resolvents.
Let your reduced cubic be $y^{3}-3py-2q=0$.
(Will substitute $p=\frac{1}{9}, q=-\frac{25}{54}$ later.)
Now $uv=p$ and $2q=u^{3}+v^{3}$.
$\therefore 2q=u^{3}+\left( \frac{p}{u} \right)^{3}$
$\implies u^{6}-2qu^{3}-p^{3}=0$
$\implies u^{3}=q\pm \sqrt{q^{2}-p^{3}}$
$\implies u=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}$
$\displaystyle{
  \therefore
v=\frac{p}{\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}}
 =\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}
}$
$\therefore y=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}+\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}$.
Note that $u, v$ are conjugates and symmetrical in roles.
By keeping the upper case,
$y=\sqrt[3]{q+\sqrt{q^{2}-p^{3}}}+\sqrt[3]{q-\sqrt{q^{2}-p^{3}}}$ is one of the root.
The other two roots are $\omega u+\omega^{2} v$ and $\omega^{2} u+\omega v$.
In this case,
$u+v=-\frac{1}{3} \left( \sqrt[3]{\frac{25+3\sqrt{69}}{2}}+\sqrt[3]{\frac{25-3\sqrt{69}}{2}} \right) $.
Careful manipulation in $\omega$ enables us to find the rest (complex roots).
P.S.: In case of the discriminant $\Delta = q^{2}-p^{3} <0$, there are 3 distinct real roots involving combination of $\cos(\frac{1}{3} \cos^{-1} (.))$ and not discussed here.
A: In theory, the answer to your question of finding the other two roots of a cubic equation knowing one of them is simply: it is enough to divide the cubic polynomial $ax^3+bx^2+cx+d$ by the linear factor $(x-x_0)$, where $x_0$ is the root you know, so you get a quadratic trinomial you can solve (this goes without saying that the value $x_0$ must be exact and not approximate)
However, in practice this is not always easily feasible and when $x_0$ has a complicated expression the above mentioned division could be hard 
In the particular case of your equation, changing  $3y$ by $x$ you have the trinomial 
$x^3-3x+25$  and the real root $x_1$ 
$$x_1=-{\sqrt[3]{\frac{2}{25-3\sqrt{69}}}}-{\sqrt[3]{\frac{25-3\sqrt{69}}{2}}}$$ 
Put by confort $$M=\frac{2}{25-3\sqrt{69}}\space\text{so}\space x_1=-\left(\sqrt[3]M+\sqrt[3]{\frac1M}\right)$$ the quotient above mentioned is equal to 
$$x^2-\left(\sqrt[3]M+\sqrt[3]{\frac 1M}\right)x+\left(\left(\sqrt[3]M+\sqrt[3]{\frac 1M}\right)^2-3\right)$$ Hence, solving this quadratic gives
$$2x_{2,3}=\sqrt[3]M+\sqrt[3]{\frac 1M}\pm\sqrt{12-3\left(\sqrt[3]M+\sqrt[3]\frac 1M\right)^2}$$
This gives the other two roots that are not real. But the calculation of its values, being direct, can be tedious.
