How do you factor # 27y^3+64#?

1 Answer
May 2, 2016

Answer:

#27y^3+64=(3y+4)(9y^2-12y+16)#

Explanation:

Note that both #27y^3 = (3y)^3# and #64=4^3# are perfect cubes.

So it is natural to use the sum of cubes identity:

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

with #a=3y# and #b=4# as follows:

#27y^3+64#

#=(3y)^3+4^3#

#=(3y+4)((3y)^2-(3y)(4)+4^2)#

#=(3y+4)(9y^2-12y+16)#