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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

my answer to this is

$\displaystyle-\left[\frac{2(b-2c)^2b}{9(a+3b)}\right]^m$

i used some factorization of some quantity to arrive to this answer. but i'm not sure how did that technique works.

for example in the quantity $(2b-4c)^{2+m}$ i factored out $2^{2+m}$ to that quantity to have $2^{2+m}(b-2c)^{2+m}$ but i don't know what theorem made this step valid can you tell me why this factorization works?

thanks!