LaTeX-examples/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.2.tex

The second solution of $x^3+\alpha x + \beta=0$ is
\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
     -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]

We will verify it in multiple steps. First, calculate $x^3$:
\begin{align}
    x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
           \underbrace{- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}}\\
         &\hphantom{{}=}+ \underbrace{3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}}
           \underbrace{- \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}}}
\end{align}

Now simplify the summands of $x^3$:
\begin{align}
    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} &=
    \left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3\\
    &= \frac{-8\alpha^3}{12 t^3}\\
    &= \frac{-2 \alpha^3}{3 t^3}\\
    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} &=- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)\\
    &= \frac{-3\alpha^2(-2(1-i\sqrt{3}))(1-i\sqrt{3})t}{t^2 \sqrt[3]{2^4 \cdot 3^2} \cdot 2 \sqrt[3]{2 \cdot 3^2}}\\
    &= \frac{6\alpha^2 t (-2 (1+i \sqrt{3}))}{12 t^2 \sqrt[3]{12}}\\
    &= \frac{- \alpha^2 (1+i\sqrt{3})}{t\sqrt[3]{12}}\\
    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} &= 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2\\
    &= \frac{3\alpha t (1+i\sqrt{3})(-2(1+i\sqrt{3}))}{4 \cdot \sqrt[3]{12 \cdot 18^2}}\\
    &= \frac{-\alpha t (-2 (1 - i \sqrt{3}))}{2 \sqrt[3]{12 \cdot 4 \cdot 3}}\\
    &= \frac{\alpha t (1-i\sqrt{3})}{\sqrt[3]{2^4 \cdot 3^2}}\\
    &= \frac{\alpha t (1-i\sqrt{3}}{2 \sqrt[3]{18}}\\
    \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}} &= - \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3\\
    &=- \frac{(-8) t^3}{8 \cdot 18}\\
    &= \frac{t^3}{18}
\end{align}

Now get back to the original equation:
\begin{align}
    0 &\stackrel{!}{=} x^3 + \alpha x + \beta \\
       &= \left (\frac{-2 \alpha^3}{3 t^3}
        + \color{red}\frac{-\alpha^2(1+\sqrt{3}i)}{t\sqrt[3]{12}}\color{black}
        + \color{blue}\frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\color{black}
        + \frac{t^3}{18} \right )\\
    &\hphantom{{}=} + \alpha \left (\color{red}\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \color{black}
     \color{blue}-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}} \color{black} \right ) + \beta\\
    &= \frac{-2 \alpha^3}{3 t^3}
    + \frac{t^3}{18}
    + \beta\\
    &= \frac{-12 \alpha^3 + t^6+18 t^3 \beta}{18t^3}
\end{align}

Now continue with only the numerator
\begin{align}
    0 &\stackrel{!}{=}
    - 12 \alpha^3
    + (\sqrt{3(4 \alpha^3 + 27 \beta^2)}-9\beta)^2
    + 18 (\sqrt{3(4 \alpha^3 + 27 \beta^2)} - 9 \beta) \beta\\
    &=
    \color{red}- 12 \alpha^3 \color{black}+
    \left (
        3(\color{red}4 \alpha^3\color{black} + \color{blue}27 \beta^2 \color{black})
        \color{orange}- 2 \cdot \sqrt{3(4 \alpha^3 + 27 \beta^2)} \cdot 9\beta\color{black}
        + \color{blue}81 \beta^2\color{black}
    \right )\\
    &\hphantom{{}=}+ 18 \beta (\color{orange}\sqrt{3(4 \alpha^3 + 27 \beta^2)}\color{black} \color{blue}- 9 \beta\color{black}) \\
    &= 81 \beta^2 + 81 \beta^2 - 2 \cdot 81 \beta^2\\
    &= 0
\end{align}