Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$

Question

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??

Answer 1

Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have $$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$

Answer 2

$$x^3=4+3\left ( \sqrt[3]{2-\sqrt{3}} \right )\left ( \sqrt[3]{2+\sqrt{3}} \right )\left ( \sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}} \right )$$ $$=4+3\sqrt[3]{2^2-3}(x)=4+3x$$ $$x^3=4+3x$$ $$x^3-3x-4=0$$

Answer 3

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: $$ q(x) = x^3-3x-4.$$

Answer 4

$$x^3=(r+s)^3=r^3+3r^2s+3rs^2+s^3=r^3+s^3+3rs(r+s)=4+3x,$$ because $r^3+s^3=4$ and $rs=1$.

Answer 5

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

$$x^3-3x-4=0$$

Answer 6

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 $p=-3$.

Answer 7

$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\\ \implies x^3=2-\color{red}{\sqrt3}+2+\color{red}{\sqrt3}+3(2-\sqrt3)(2+\sqrt3)(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}})\\ \implies x^3=4+3.1.\color{red}{x}\\ \implies x^3-4-3x=0$

Answer 8

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 $b = 0$.

$$\begin{cases} a = 1 \\ b = 0 \\ 2 = -\frac{d}{2} \\ 3 = \left( \frac{c}{3} \right)^3 + \left( \frac{d}{2} \right)^2 \\ \end{cases}$$

$$\begin{cases} a = 1 \\ b = 0 \\ 3 = \left( \frac{c}{3} \right)^3 + 4 \\ d = -4 \\ \end{cases}$$

$$\begin{cases} a = 1 \\ b = 0 \\ c = -3 \\ d = -4 \\ \end{cases}$$

$$x^3 - 3x - 4 = 0$$