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} 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 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 \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 + 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} \goodbreak Now back to the original equation: \begin{align} 0 &\stackrel{!}{=} x^3 + \alpha x + \beta\\ &= \frac{-96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3}) + 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6}{12^2 t^3}\\ &\hphantom{{}=}+\alpha \left (\sqrt[3]{12} \cdot \frac{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}{12t} \right ) + \beta \end{align} \todo[inline]{the calculation above seems to be wrong / too long. Next try} When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance} you get:\footnote{Remember, that $(1+i\sqrt{3})^2 = -2 (1-i \sqrt{3})$ and $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$ and $(1 \pm i \sqrt{3})^3 = -8$} \begin{align} 0 &\stackrel{!}{=} \left( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3 + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{(1-i \sqrt{3})^3 \alpha^3}{12 \cdot t^3} - 3 \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \right )^2 \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} + 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \left (\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^2\\ &\hphantom{{}=} + \frac{(1+i\sqrt{3})^3 t^3}{2^3 \cdot 18} + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{-8 \alpha^3}{12t^3} - 3 \frac{-2(1+i \sqrt{3}) \alpha^2}{\sqrt[3]{12^2} t^2} \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} + 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \frac{-2(1-i\sqrt{3}) t^2}{4\sqrt[3]{18^2}}\\ &\hphantom{{}=} + \frac{-8 t^3}{2^3 \cdot 18} + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{-2 \alpha^3}{3t^3} + \frac{6 \alpha^2 t (-2)(1-i \sqrt{3})}{(\sqrt[3]{12^2} t^2)(2\sqrt[3]{18})} + \frac{12 \alpha t^2 (1+i \sqrt{3})}{(\sqrt[3]{12} \cdot t)(4\sqrt[3]{18^2)}}\\ &\hphantom{{}=} + \frac{- t^3}{18} + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{-2 \alpha^3}{3t^3} + \frac{-6 \alpha^2 (1-i \sqrt{3})}{6 \sqrt[3]{12} t} + \frac{3 \alpha t (1+i \sqrt{3})}{6\sqrt[3]{18}} + \frac{- t^3}{18}\\ &\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{-2 \alpha^3}{3t^3} + \frac{-\alpha^2 (1-i \sqrt{3})}{\sqrt[3]{12} t} + \frac{\alpha t (1+i \sqrt{3})}{2\sqrt[3]{18}} + \frac{- t^3}{18}\\ &\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &= \frac{12 \cdot (-2 \alpha^3) +(6 \sqrt[3]{18}t^2)(-\alpha^2 (1-i \sqrt{3}))+ (3 \sqrt[3]{12})(\alpha t (1+i \sqrt{3})) + (2t^3)(- t^3)}{36t^3}\\ &\hphantom{{}=}+ \frac{(6 \sqrt[3]{18})((1-i \sqrt{3}) \alpha) - (3 \sqrt[3]{12})((1+i\sqrt{3}) t) + 36t^3 \beta}{36t^3}\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{align} \goodbreak Now calculate only the numerator: \begin{align} 0 &\stackrel{!}{=} -12 \alpha^3 - 6 \sqrt[3]{18} t^2 \alpha^2 (1 - i \sqrt{3}) + 3 \sqrt[3]{12} \alpha t (1+i\sqrt{3}) - 2t^6\\ &\hphantom{{}=} + 6\sqrt[3]{18} \alpha (1- i \sqrt{3}) - 3 \sqrt[3]{12} t (1+i \sqrt{3}) + 36 t^3 \beta \end{align}