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 Sep 29, 2018
 #1
avatar+107446 
+2

It would be good if another mathematician checks this.  I think it is ok. 

Heureka taught me just a little while ago.

 

 

Find the coefficient of   \(a^4b^4\)   in the expansion of   \((\frac{a^2}{3}-2b)^6\)

 

what you need first is the coefficient of      \((\frac{a^2}{3})^2\;(-2b)^4\)

 

it will be      \(\frac{6!}{2!4!}=15\) 

 

so we have       \(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\)

 

 

\(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\\ =15\cdot (\frac{a^4}{9})\;\cdot 16b^4\\ =5\cdot (\frac{a^4}{3})\;\cdot 16b^4\\ =\frac{80}{3} a^4b^4\\ \)

 

so the coefficient is      \(\frac{80}{3}\)

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 Sep 29, 2018
 #2
avatar+2345 
+1

This looks good to me!

 Sep 29, 2018
 #3
avatar+107446 
0

Thank you   laugh

Melody  Sep 30, 2018

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