(m+(mn^2) ^1/3 + (m^2 n)^1/3)/ m-n * (1- (n^1/3)/(m^1/3))

asked by guest
on Mar 29, 2025 at 6:31 am



You asked:

Evaluate the expression: m+(mn2)13+(m2n)13m−n(1−n13m13)\frac{m + \frac{{\left( m {n}^{2} \right)}^{1}}{3} + \frac{{\left( {m}^{2} n \right)}^{1}}{3}}{m} - n \left(1 - \frac{\frac{{n}^{1}}{3}}{\frac{{m}^{1}}{3}}\right)

MathBot Answer:

Evaluated



m+(mn2)13+(m2n)13m−n(1−n13m13)=−n(1−nm)+m2n3+n2m3+mm\displaystyle \frac{m + \frac{{\left( m {n}^{2} \right)}^{1}}{3} + \frac{{\left( {m}^{2} n \right)}^{1}}{3}}{m} - n \left(1 - \frac{\frac{{n}^{1}}{3}}{\frac{{m}^{1}}{3}}\right) = - n \left(1 - \frac{n}{m}\right) + \frac{\frac{m^{2} n}{3} + \frac{n^{2} m}{3} + m}{m}

Expanded

m+(mn2)13+(m2n)13m−n(1−n13m13)=mn3+n23−n+1+n2m\frac{m + \frac{{\left( m {n}^{2} \right)}^{1}}{3} + \frac{{\left( {m}^{2} n \right)}^{1}}{3}}{m} - n \left(1 - \frac{\frac{{n}^{1}}{3}}{\frac{{m}^{1}}{3}}\right) = \frac{m n}{3} + \frac{n^{2}}{3} - n + 1 + \frac{n^{2}}{m}

Factored

m+(mn2)13+(m2n)13m−n(1−n13m13)=m2n+n2m−3mn+3m+3n23m\frac{m + \frac{{\left( m {n}^{2} \right)}^{1}}{3} + \frac{{\left( {m}^{2} n \right)}^{1}}{3}}{m} - n \left(1 - \frac{\frac{{n}^{1}}{3}}{\frac{{m}^{1}}{3}}\right) = \frac{m^{2} n + n^{2} m - 3 m n + 3 m + 3 n^{2}}{3 m}

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