How do you condense #((3a^3b^3)/(a-b))/((4ab)/(b-a))#?

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Answer 1 · 2017-02-20T08:04:47

#((3a^3b^3)/(a-b))/((4ab)/(b-a))=-(3a^2b^2)/4#

Note thar #(b-a)--(a-b)# and we are going to use this here.

Now #((3a^3b^3)/(a-b))/((4ab)/(b-a))#

= #(3a^3b^3)/((a-b))xx((b-a))/(4ab)#

= #(3a^3b^3)/((a-b))xx-((a-b))/(4ab)#

= #(3a^3b^3)/(cancel((a-b)))xx-(cancel((a-b)))/(4ab)#

= #- (3xxaxxaxxcancelaxxcancelbxxbxxb)/(4xxcancelaxxcancelb)#

= #-(3a^2b^2)/4#