Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is 
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??

The general solution to a cubic:
$$\begin{align}
r & =  \sqrt[3]{
\underbrace{  \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) }_{p} 
+ \sqrt[2]{
    \left(\underbrace{  \frac{c}{3a} - \frac{b^2}{9a^2} }_{q} \right)^3   + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
  }
} \\
& +  \sqrt[3]{ 
\left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) 
- \sqrt[2]{
     \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3  + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2 
  }
} \\
& - \underbrace{ \frac{b}{3a} }_{r}
\end{align}$$
The problem gives $p = 2$, $q^3 + p^2= 3$, $r = 0$.  
$$\begin{cases}
0 = \frac{b}{3a} \\
2 = \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \\
3 = \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3  + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2  \\
\end{cases}$$
Since polynomials have the same roots up to a scaling factor (that is, $f(x)$ has the same roots as $k~f(x)$), we can choose $a = 1$.  And solve and get $b = 0$.
$$\begin{cases}
a = 1 \\
b = 0 \\
2 = -\frac{d}{2} \\
3 = \left( \frac{c}{3} \right)^3  + \left( \frac{d}{2} \right)^2  \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
3 = \left( \frac{c}{3} \right)^3  + 4  \\
d = -4 \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
c = -3 \\
d = -4 \\
\end{cases}$$
$$x^3 - 3x - 4 = 0$$
