How do you factor #8a^3+27(x+y)^3#?

2 Answers
Apr 11, 2015

The answer is:

#(2a+3x+3y)(4a^2-6ax-6ay+9x^2+18xy+9y^2)#

If you write:

#8a^3+27(x+y)^3=(2a)^3+[3(x+y)]^3=(1)#

This is a sum of two cubic, and you can use this rule:

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

So:

#(1)=[2a+3(x+y)][(2a)^2-(2a)*3(x+y)+3^2(x+y)^2]=#

#=(2a+3x+3y)(4a^2-6ax-6ay+9x^2+18xy+9y^2)#.

Apr 11, 2015

Temporarily simplify be replacing
#8a^3 = (2a)^3# with #p^3#
and
#27(x+y)^3 = (3(x+y))^3# with #q^3#

So #8a^3+27(x+y)^3#
becomes #p^3+q^3#
which has factors
#(p+q)(p^2-pq+q^2)#

Restoring the original values, this becomes
#(2a + 3(x+y))*(4a^2 - 6a(x+y) +9(x+y)^2)#

It might be tempting to attempt to further factor the right-most term but a quick check using the formula for roots
#(-b+-sqr(b^2-4ac))/2a# reveals that no Real roots are possible
[be careful not to confuse the #a# in this general formula with the #a# in the example expression]