So remember some exponentional laws

and

so
4 to the 6th root=4^(1/6)
4 to the 3rd root=4^(1/3)
so

and
![4^{-1/3}= \frac{1}{4^{1/3}}= \frac{1}{ \sqrt[3]{4} }](https://tex.z-dn.net/?f=%204%5E%7B-1%2F3%7D%3D%20%5Cfrac%7B1%7D%7B4%5E%7B1%2F3%7D%7D%3D%20%5Cfrac%7B1%7D%7B%20%5Csqrt%5B3%5D%7B4%7D%20%7D%20%20%20)
the answer is
Hope I helped with your problem
7 and 21 can cancel leaving a 3 on the bottom. one m can cancel from the bottom and the top so that leaves you with m^3 n^4/3
now one denominator is x^2 + 8x + 15
if we factorise then (x +3)(x +5)
now in expressions which have like x+3 as denominator ...you multiply and divide them by x+5 and once u get denominator equal you can add numerator easily