Question Number 185198 by ajfour last updated on 18/Jan/23
Commented by ajfour last updated on 18/Jan/23
$${If}\:\:\:{z}={r}+{xi}+{yj} \\ $$$${z}^{\mathrm{2}} ={r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{r}\left({xi}+{yj}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:+{xy}\left({ij}+{ji}\right) \\ $$$${let}\:\:\:{ji}={i}\:\:\:{and}\:\:{ij}={j} \\ $$$$\Rightarrow\:\:{z}^{\mathrm{2}} ={r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{x}\left(\mathrm{2}{r}+{y}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{y}\left(\mathrm{2}{r}+{x}\right){j} \\ $$$${z}^{\mathrm{3}} ={r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+{rx}\left(\mathrm{2}{x}+{y}\right){i} \\ $$$$\:\:\:\:\:+{ry}\left(\mathrm{2}{r}+{x}\right){j}+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){i} \\ $$$$\:\:\:\:−{x}^{\mathrm{2}} \left(\mathrm{2}{r}+{y}\right)+{xy}\left(\mathrm{2}{r}+{x}\right){j} \\ $$$$\:\:\:+{y}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){j}+{xy}\left(\mathrm{2}{r}+{y}\right){i} \\ $$$$\:\:\:\:−{y}^{\mathrm{2}} \left(\mathrm{2}{r}+{x}\right) \\ $$$$\:{z}^{\mathrm{3}} \:={r}^{\mathrm{3}} −{r}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \right)−{xy}\left({x}+{y}\right) \\ $$$$+{i}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}{ry}\right){x}+{j}\left({r}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{2}{rx}\right){y} \\ $$$$ \\ $$$$\bigstar \\ $$
Commented by ajfour last updated on 18/Jan/23
$${Now}\:{if}\:{p}^{\mathrm{3}} −{p}−{c}=\mathrm{0}\:\:{with}\:{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${then}\:\:{p}=\left(\frac{{c}}{\mathrm{2}}+{i}\sqrt{\frac{\mathrm{1}}{\mathrm{27}}−\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\frac{{c}}{\mathrm{2}}−{j}\sqrt{\frac{\mathrm{1}}{\mathrm{27}}−\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${for}\:\:{c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${p}=\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{{i}}{\mathrm{6}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{1}}{\mathrm{6}}−\frac{{j}}{\mathrm{6}\sqrt{\mathrm{3}}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${p}\sqrt{\mathrm{3}}=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{{j}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:=\left({w}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} +\left({z}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$$${let}\:\:\:{w}={r}_{\mathrm{1}} +{xi}+{yj} \\ $$$$\:\:\:\:\:\:\:\:\:\:{z}={r}_{\mathrm{2}} −{xi}−{yj} \\ $$$${p}\sqrt{\mathrm{3}}={w}+{z}={r}_{\mathrm{1}} +{r}_{\mathrm{2}} \\ $$$$ \\ $$