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# Sorry Melody - I wrote the Question Down Wrong

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Melody the question I wrote was wrong. It was actually:

Write $$\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3$$as a power of 2.

Jan 18, 2022

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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,

Jan 18, 2022
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MathyGoo13  Jan 19, 2022
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$$\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\ =\frac{4}{27}\left( \left[ 2^{\frac{3+m}{3}} \right] + \left[ 2^{\frac{m}{3}} \right] \right)^3\\ =\frac{4}{27}\left(\left[ 2^{\frac{3+m}{3}} \right]^3 + 3*\left[ 2^{\frac{3+m}{3}} \right]^2 \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] ^2 + \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+2m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{2m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+3m}{3}} \right] + 3*\left[ 2^{\frac{3+3m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{2+m} \right] + 3*\left[ 2^{1+m} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4*2^m}{27}\left(\left[ 2^{3} \right] + 3*\left[ 2^{2} \right] + 3*\left[ 2 \right] + \left[ 1 \right] \right)\\ =\frac{4*2^m}{27}\left(8+ 12 + 6 + 1\right)\\ =\frac{4*2^m}{27}\left(27\right)\\ =2^{(m+2)}$$

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)}

Jan 19, 2022
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Normally I would not have answered this on athis new thread, but the original unintentioned question was a completely different question.

Jan 19, 2022