How do you factor # 27y^3+64#?
1 Answer
May 2, 2016
Explanation:
Note that both
So it is natural to use the sum of cubes identity:
#a^3+b^3=(a+b)(a^2-ab+b^2)#
with
#27y^3+64#
#=(3y)^3+4^3#
#=(3y+4)((3y)^2-(3y)(4)+4^2)#
#=(3y+4)(9y^2-12y+16)#