Find the roots of the polynomial? (Cardano's Method) $y^3-\frac7{12}y-\frac7{216}$
This is part of Cardano's method, so I've gotten my first root to be:
$y_1=\sqrt[3]{\frac7{432}+i\sqrt{\frac{49}{6912}}}+\sqrt[3]{\frac7{432}-i\sqrt{\frac{49}{6912}}}$
I am at a loss on how to evaluate this. I know this is one of three real roots, but I don't know what to do now.
 A: Well, you can write
$${7\over 432}+ i\sqrt{49\over6912}={7\sqrt{7}\over216}\,e^{i\arctan\left(3\sqrt{3}\right)}=\left({\sqrt{7}\over6}\right)^3\,e^{i\arctan\left(3\sqrt{3}\right)}\\
{7\over 432}- i\sqrt{49\over6912}={7\sqrt{7}\over216}\,e^{-i\arctan\left(3\sqrt{3}\right)}=\left({\sqrt{7}\over6}\right)^3\,e^{-i\arctan\left(3\sqrt{3}\right)}$$
And go from there with determinations of $\sqrt[3]{z}$, if it helps you.
The formula I used is $\forall a,b\in \mathbb{R}\ s.t. a\neq0,\ a+ib=\sqrt{a^2+b^2}\,e^{i\arctan(b/a)}$
A: The analytical method for the irreducible case compares
$$
y^3−\frac{7}{12}y=\frac{7}{216}
$$
with $y=r\cosϕ$, 
$$
y^3=r^3\frac{(e^{iϕ}+e^{-iϕ})^3}{8}=\frac{r^3}4(\cos(3ϕ)+3\cosϕ)
\\
\iff\\ y^3-\frac34 r^2y=\frac{r^3}4\cos(3ϕ)
$$
to read off $r=\frac{\sqrt{7}}3$ and
$$
\frac{7\sqrt{7}}{4·3^3}\cos(3ϕ)=\frac{7}{2^3·3^3}
\iff 
ϕ=\frac13·\arccos\frac1{2\sqrt7}+k\frac{2\pi}3, \quad k=-1,0,1
$$
A: Doing what G. Sassatelli suggested in her/his answer (all credit being for her/him) and being a little patient, you could arrive to three real roots (this is proven using Cardano) which are $$y_1=\frac{1}{3} \sqrt{7} \cos \left(\frac{1}{3} \tan ^{-1}\left(3 \sqrt{3}\right)\right)$$ $$y_2=-\frac{1}{2} \sqrt{\frac{7}{3}} \sin \left(\frac{1}{3} \tan ^{-1}\left(3
   \sqrt{3}\right)\right)-\frac{1}{6} \sqrt{7} \cos \left(\frac{1}{3} \tan
   ^{-1}\left(3 \sqrt{3}\right)\right)$$ $$y_3=\frac{1}{2} \sqrt{\frac{7}{3}} \sin \left(\frac{1}{3} \tan ^{-1}\left(3
   \sqrt{3}\right)\right)-\frac{1}{6} \sqrt{7} \cos \left(\frac{1}{3} \tan
   ^{-1}\left(3 \sqrt{3}\right)\right)$$
A: One can proceed as described above and find the three roots:
$\left\{\frac{1}{6} \left(\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(1+3 i
   \sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(1+3 i
   \sqrt{3}\right)}\right),-\frac{7^{2/3} \left(1+i \sqrt{3}\right)}{6\ 2^{2/3}
   \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2}
   \left(1+3 i \sqrt{3}\right)},-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{6\ 2^{2/3}
   \sqrt[3]{1+3 i \sqrt{3}}}-\frac{1}{12} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2}
   \left(1+3 i \sqrt{3}\right)}\right\}$
