How do you factor #(a+2b)^3 - (a-2b)^3#?

1 Answer
Apr 15, 2015

Remembering that a difference of cubes can be factored:

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

then:

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

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

#=(a+2b-a+2b)(a^2+4ab+4b^2+a^2-4b^2+a^2-4ab+4b^2]=#

#=4b(3a^2+4b^2)#.

But, in this case, there is a faster and easier way:

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

#=a^3+6a^2b+12ab^2+8b^3-a^3+6a^2b-12ab^2+8b^3=#

#=12a^2b+16b^3=4b(3a^2+4b^2)#.