Sorry Melody - I wrote the Question Down Wrong

Thanks Mathy.

Now it comes out perfectly well.

If you still need help with it please put this correct version, along with the explantatory note,  on the original thread.   

I can help you over there is you still want me too,

\(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\ =\frac{4}{27}\left( \left[ 2^{\frac{3+m}{3}} \right] + \left[ 2^{\frac{m}{3}} \right] \right)^3\\ =\frac{4}{27}\left(\left[ 2^{\frac{3+m}{3}} \right]^3 + 3*\left[ 2^{\frac{3+m}{3}} \right]^2 \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] ^2 + \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+2m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{2m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+3m}{3}} \right] + 3*\left[ 2^{\frac{3+3m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{2+m} \right] + 3*\left[ 2^{1+m} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4*2^m}{27}\left(\left[ 2^{3} \right] + 3*\left[ 2^{2} \right] + 3*\left[ 2 \right] + \left[ 1 \right] \right)\\ =\frac{4*2^m}{27}\left(8+ 12 + 6 + 1\right)\\ =\frac{4*2^m}{27}\left(27\right)\\ =2^{(m+2)} \)

LaTex

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

=\frac{4}{27}\left(     \left[  2^{\frac{3+m}{3}} \right]    +    \left[ 2^{\frac{m}{3}} \right]    \right)^3\\
=\frac{4}{27}\left(\left[  2^{\frac{3+m}{3}} \right]^3 
+ 3*\left[  2^{\frac{3+m}{3}} \right]^2   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]  ^2  +  \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+2m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{2m}{3}} \right]   +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+3m}{3}} \right]      
+ 3*\left[  2^{\frac{3+3m}{3}} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{2+m} \right]      
+ 3*\left[  2^{1+m} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4*2^m}{27}\left(\left[  2^{3} \right] 
+ 3*\left[  2^{2} \right]      
+ 3*\left[  2 \right]     +  \left[ 1 \right]  \right)\\

=\frac{4*2^m}{27}\left(8+  12 + 6    + 1\right)\\
=\frac{4*2^m}{27}\left(27\right)\\
=2^{(m+2)}

Normally I would not have answered this on athis new thread, but the original unintentioned question was a completely different question.