Question

Simplify. (m+n/7)^3 (m-n/7)

Answer 1

Solution:-

given by:-

${(m + \frac{n}{7}) }^{3} (m - \frac{n}{7} ) \\(m + \frac{n}{7} )(m - \frac{m}{7} ) {(m + \frac{n}{7} )}^{2} \\ ({m}^{2} - \frac{ {n}^{2} }{49} )( {m}^{2} + \frac{ {n}^{2} }{49} + \frac{2mn}{7}) \\ {m}^{4} + \frac{ {m}^{2} {n}^{2} }{49} + \frac{2 {m}^{3}n }{7} - \frac{ {m}^{2} {n}^{2} }{49} - \frac{ {n}^{4} }{2401} - \frac{2m {n}^{3} }{7} \\ {m}^{4} - \frac{ {n}^{4} }{2401} + \frac{2 {m}^{3}n }{7} - \frac{2m {n}^{3} }{7} \\ \\ i \: hope \: its \: is \: helpfull \: for \: u$

Answer 2

To simplify: $\left(m+\frac{n}{7}\right)^3\:\left(m-\frac{n}{7}\right)$

Consider $\left(m+\frac{n}{7}\right)^3\:\left(m-\frac{n}{7}\right)$

$=\left(m+\frac{n}{7}\right)^3m-\left(m+\frac{n}{7}\right)^3\frac{n}{7}\ \ \ [\text{By distributive law}:a\left(b-c\right)=ab-ac]\\\\ =m\left(m+\frac{n}{7}\right)^3-\frac{n}{7}\left(m+\frac{n}{7}\right)^3\\\\ =m\left(m^3+(\frac{n}{7})^3+3m^2(\dfrac{n}7)+3m(\dfrac{n}7)^2\right)-\frac{n}{7}\left((m^3+(\frac{n}{7})^3+3m^2(\dfrac{n}7)+3m(\dfrac{n}7)^2\right)\ \ \ [\because \ \ (a+b)^3=a^3+b^3+3a^2b+3ab^2]$

$=m^4+\dfrac{mn^3}{343}+\dfrac{3m^3n}{7}+\dfrac{3m^2n^2}{49}-\dfrac{nm^3}{7}-\dfrac{n^4}{2401}-\dfrac{3m^2n^2}{49}-\dfrac{3mn^3}{343}\\\\=m^4-\frac{n^4}{2401}-\frac{2mn^3}{343}+\frac{2m^3n}{7}$

Hence, the simplified expression would be $m^4-\frac{n^4}{2401}-\frac{2mn^3}{343}+\frac{2m^3n}{7}$.