How do you factor #5u^3-40(x+y)^3#?

1 Answer
May 15, 2016

Answer:

#5u^3-40(x+y)^3=5(u-2x-2y)(u^2+2ux+2uy+4x^2+8xy+4y^2)#

Explanation:

We will use the difference of cubes identity:

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

with #a=u# and #b=2(x+y)#

First separate out the common scalar factor #5#:

#5u^3-40(x+y)^3#

#=5(u^3-8(x+y)^3)#

#=5(u^3-(2(x+y))^3)#

#=5(u-2(x+y))(u^2+u(2(x+y))+(2(x+y))^2)#

#=5(u-2x-2y)(u^2+2ux+2uy+4x^2+8xy+4y^2)#