+0  
 
0
234
1
avatar

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

 Jan 18, 2022
 #1
avatar
0

AMOGUS!

 Nov 2, 2022

3 Online Users