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documents/math-minimal-distance-to-cubic-function/quadratic-case-2.3.tex
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\todo[inline]{calculate...}
One solution 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, get $x^3$:
\begin{align}
x^3 &= \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} - \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3\\
&= \left (\frac{(2\sqrt[3]{18})(1-i \sqrt{3}) \alpha - (\sqrt[3]{12} \cdot t)(1+i\sqrt{3}) t}{\sqrt[3]{12} \cdot t \cdot 2 \cdot \sqrt[3]{18}} \right)^3\\
&= \left (\frac{2\sqrt[3]{18}a \alpha (1-i \sqrt{3}) - \sqrt[3]{12} \cdot t^2(1+i\sqrt{3})}{2t \sqrt[3]{2^3 \cdot 3^3}} \right )^3\\
&= 12 \cdot \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}^{\text{numerator}}}{12t} \bigg )^3
&= \left (\frac{(2\sqrt[3]{18})(1-i \sqrt{3}) \alpha - (\sqrt[3]{12} \cdot t)(1+i\sqrt{3}) t}{\sqrt[3]{12} t \cdot 2 \sqrt[3]{18}} \right)^3\\
&= \left (\frac{2\sqrt[3]{18}\alpha (1-i \sqrt{3}) - \sqrt[3]{12} t^2(1+i\sqrt{3})}{2t \cdot 6} \right )^3\\
&= \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2 (1+i\sqrt{3})}^{\text{numerator}}}{\sqrt[3]{12^2} t} \bigg )^3
\end{align}
Now calculate the numerator$^3$:
Now calculate numerator$^3$:
\begin{align}
\left (\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3}) \right )^3 &=
12 \alpha^3 (1-i\sqrt{3})^3 \\
&\hphantom{{}=}- 3 \sqrt[3]{12^2} \alpha^2(1-i\sqrt{3})^2 (t^2(1+i \sqrt{3}))\\
&\hphantom{{}=}+ 3 \sqrt[3]{12\hphantom{^2}} \alpha\hphantom{^2} (1-i\sqrt{3}) t^4 (1+i\sqrt{3})^2 - t^6 (1+i\sqrt{3})^3\\
&= 12 \alpha^3 \cdot (-8) \\
&\hphantom{{}=}- 3 \cdot 2 \sqrt[3]{18} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\
&\hphantom{{}=}- 3 \sqrt[3]{12^2} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\
&\hphantom{{}=}+ 3 \sqrt[3]{12} \alpha (1-i\sqrt{3}) t^4 (-2(1-i\sqrt{3})) - t^6 (-8)\\
&= -96 \alpha^3 + 12 \sqrt[3]{18} \alpha^2 t^2 (1+i \sqrt{3})^2\\
&\hphantom{{}=}- 6 \sqrt[3]{12} \alpha (1-i\sqrt{3})^2 t^4 +8 t^6\\
&= -96 \alpha^3 + 6 \sqrt[3]{12^2} \alpha^2 t^2 (1+i \sqrt{3})^2\\
&\hphantom{{}=}- 6 \sqrt[3]{12} \alpha t^4 (1-i\sqrt{3})^2 +8 t^6\\
&= -96 \alpha^3 - 12 \sqrt[3]{12^2} \alpha^2 t^2 (1-i \sqrt{3})\\
&\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6\\
&= -96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3})\\
&\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6
\end{align}