Question Number 171346 by Mr.D.N. last updated on 13/Jun/22
$$\:\:\mathrm{solve}: \\ $$$$\:\:\:\frac{\mathrm{m}+\left(\mathrm{mn}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{m}^{\mathrm{2}} \mathrm{n}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{m}−\mathrm{n}}\:×\:\frac{\left(\mathrm{m}^{\mathrm{1}/\mathrm{3}} −\mathrm{n}^{\mathrm{1}/\mathrm{3}} \right)}{\mathrm{m}^{\mathrm{1}/\mathrm{3}} } \\ $$
Commented by infinityaction last updated on 13/Jun/22
$$\frac{\cancel{{m}^{\mathrm{1}/\mathrm{3}} }\left\{{m}^{\mathrm{2}/\mathrm{3}} +{n}^{\mathrm{2}/\mathrm{3}} +\left({mn}\right)^{\mathrm{1}/\mathrm{3}} \right\}}{{m}−{n}}×\frac{{m}^{\mathrm{1}/\mathrm{3}} −{n}^{\mathrm{1}/\mathrm{3}} }{\cancel{{m}^{\mathrm{1}/\mathrm{3}} }} \\ $$$$\:\frac{\left\{{m}^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{3}} −\left\{{n}^{\mathrm{1}/\mathrm{3}} \right\}^{\mathrm{3}} }{{m}−{n}} \\ $$$$\:\:\:\:\cancel{\frac{{m}−{n}}{{m}−{n}}}\:\:=\:\mathrm{1} \\ $$
Commented by Mr.D.N. last updated on 13/Jun/22
$$\mathrm{thank}\:\mathrm{you}.\mathrm{great} \\ $$