How do you simplify #(a^3 - b^3)/(3a^2 + 9ab + 6b^2)*(a^2 + 2ab + b^2)/(a^2 - b^2)#?

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Answer 1 · 2015-07-19T12:16:53

Factor and cancel matching factors to find:

#(a^3-b^3)/(3a^2+9ab+6b^2)*(a^2+2ab+b^2)/(a^2-b^2)=(a^2+ab+b^2)/(3(a+2b))#

Use difference of cubes identity:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

Use difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

#(a^3-b^3)/(3a^2+9ab+6b^2)*(a^2+2ab+b^2)/(a^2-b^2)#

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

#=((a^2+ab+b^2)(a+b))/(3(a+b)(a+2b))#

#=(a^2+ab+b^2)/(3(a+2b))#