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??
 A: $$x^3=(r+s)^3=r^3+3r^2s+3rs^2+s^3=r^3+s^3+3rs(r+s)=4+3x,$$
because $r^3+s^3=4$ and $rs=1$.
A: Identifying with Cardano's formula
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}},$$
we find $q=-4$, then $p=-3$.
A: $$x^3=4+3\left ( \sqrt[3]{2-\sqrt{3}} \right )\left (  \sqrt[3]{2+\sqrt{3}} \right )\left ( \sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}} \right )$$
$$=4+3\sqrt[3]{2^2-3}(x)=4+3x$$
$$x^3=4+3x$$
$$x^3-3x-4=0$$
A: Probably it is better to notice, for first, that the product between $2-\sqrt{3}$ and $2+\sqrt{3}$ is one. After that, the minimal polynomial of $a=\sqrt[3]{2+\sqrt{3}}$ over $\mathbb{Q}$ is quite trivially:
$$ p(x)=(x^3-2)^2-3 = x^6-4x^3+1 $$
hence $a$ is a root of:
$$ x^3+x^{-3}-4 = \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right)-4$$
and $a+\frac{1}{a}$ is a root of:
$$ q(x) = x^3-3x-4.$$
A: Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have
$$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$
A: Let $a^3=2-\sqrt3,b^3=2+\sqrt3$, we have
$$a^3+b^3=4$$
$$a+b=x$$
$$ab=1$$
Consider
\begin{align}
x^3=(a+b)^3&=3ab(a+b)+a^3+b^3\\&=3x+4
\end{align}
Therefore
$$x^3-3x-4=0$$
A: $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\\
\implies x^3=2-\color{red}{\sqrt3}+2+\color{red}{\sqrt3}+3(2-\sqrt3)(2+\sqrt3)(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}})\\
\implies x^3=4+3.1.\color{red}{x}\\
\implies x^3-4-3x=0$
A: 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 $b = 0$.
$$\begin{cases}
a = 1 \\
b = 0 \\
2 = -\frac{d}{2} \\
3 = \left( \frac{c}{3} \right)^3  + \left( \frac{d}{2} \right)^2  \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
3 = \left( \frac{c}{3} \right)^3  + 4  \\
d = -4 \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
c = -3 \\
d = -4 \\
\end{cases}$$
$$x^3 - 3x - 4 = 0$$
