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20100090100
https://adamaths.com/forum/viewtopic.php?f=71&t=2166
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20100090100
Posted:
Thu Sep 28, 2017 12:44 pm
by
ada
*$-\frac{81a^5b^4}{2}+\frac{261a^5b^2}{8}+\frac{33a^5}{8}$