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

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