Manipulating Exponents

Please check the question.

$\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\ \frac{4}{27}\left((2^{\frac{3+m}{3}})^3 + 3 (2^{\frac{3+m}{3}})^2 (2^{\frac{3}{m}}) +3(2^{\frac{3+m}{3}}) (2^{\frac{3}{m}})^2 + (2^{\frac{3}{m}})^3 \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(1+\frac{m}{3})})^2 (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(2+\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{2^2}{3^3}\left((2^32^m + 3 (2^2 2^{(\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2* 2^{(\frac{m}{3})}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ $

I can keep going but this is not going to fall out nicely.

LaTex:

\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\

\frac{4}{27}\left((2^{\frac{3+m}{3}})^3  +  3 (2^{\frac{3+m}{3}})^2  (2^{\frac{3}{m}})  +3(2^{\frac{3+m}{3}})  (2^{\frac{3}{m}})^2  +   (2^{\frac{3}{m}})^3 \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(1+\frac{m}{3})})^2  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(2+\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{2^2}{3^3}\left((2^32^m  +  3 (2^2 2^{(\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2* 2^{(\frac{m}{3})})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

Here’s a neat and orderly way to transcribe this as a power of (2):

$\hspace {15em} \huge2 \large^ {\log_2\left(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\right)}\\$

GA 

--. .-

Guys im confused. 

From trials and error I found that: $2^{m+2}$ works for some reason. 

Help me with your knowledge!!