Question Number 6809 by Tawakalitu. last updated on 28/Jul/16
$${If}\:\:{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\:\:{y}^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\:\:{z}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:=\:\:\mathrm{0},\:\:{Then}\:\:{pove}\:{that}\: \\ $$$$\left({x}\:+\:{y}\:+\:{z}\right)^{\mathrm{3}} \:\:=\:\mathrm{3}{xyz} \\ $$
Commented by Yozzii last updated on 29/Jul/16
$${x}^{\mathrm{1}/\mathrm{3}} +{y}^{\mathrm{1}/\mathrm{3}} +{z}^{\mathrm{1}/\mathrm{3}} =\mathrm{0} \\ $$$${Let}\:{x}={u}^{\mathrm{3}} ,{y}={w}^{\mathrm{3}} ,{z}={v}^{\mathrm{3}} . \\ $$$$\therefore{u}+{w}+{v}=\mathrm{0} \\ $$$${Using}\:{the}\:{identity}\:{u}^{\mathrm{3}} +{w}^{\mathrm{3}} +{v}^{\mathrm{3}} =\left({u}+{w}+{v}\right)\left({u}^{\mathrm{2}} +{w}^{\mathrm{2}} +{v}^{\mathrm{2}} −{uv}−{vw}−{wu}\right)+\mathrm{3}{uwv} \\ $$$${and}\:{u}+{w}+{v}=\mathrm{0}, \\ $$$$\mathrm{3}{uvw}={u}^{\mathrm{3}} +{w}^{\mathrm{3}} +{v}^{\mathrm{3}} \\ $$$${or}\:\mathrm{3}\left({xyz}\right)^{\mathrm{1}/\mathrm{3}} ={x}+{y}+{z} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{3}} =\mathrm{27}{xyz} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{x}=\mathrm{8},{y}=\mathrm{27},\:{z}=−\mathrm{125}\:{as}\:{an}\:{example}. \\ $$$$\Rightarrow{x}^{\mathrm{1}/\mathrm{3}} +{y}^{\mathrm{1}/\mathrm{3}} +{z}^{\mathrm{1}/\mathrm{3}} =\mathrm{2}+\mathrm{3}−\mathrm{5}=\mathrm{0}. \\ $$$$\therefore\left({x}+{y}+{z}\right)^{\mathrm{3}} =−\mathrm{729000} \\ $$$${And}\:\mathrm{3}{xyz}=\mathrm{3}×\mathrm{8}×\mathrm{27}×\left(−\mathrm{125}\right)=−\mathrm{81000}\neq−\mathrm{729000} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{3}} \neq\mathrm{3}{xyz}\:{generally}\:{if}\:{x}^{\mathrm{1}/\mathrm{3}} +{y}^{\mathrm{1}/\mathrm{3}} +{z}^{\mathrm{1}/\mathrm{3}} =\mathrm{0}. \\ $$$${But},\:\mathrm{27}{xyz}=\mathrm{27}×\mathrm{8}×\mathrm{27}×\left(−\mathrm{125}\right)=−\mathrm{729000} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{3}} =\mathrm{27}{xyz}. \\ $$
Commented by Tawakalitu. last updated on 29/Jul/16
$${Thank}\:{you}\:{very}\:{much}.\: \\ $$