How do you solve #y^3 - 27 = 9y^2 - 27y#?

1 Answer
iceman
Feb 12, 2017

#y=3#

Explanation:

#y^3−27=9y^2−27y# => subtract #9y^2# from both sides:
#y^3-9y^2-27=-27y# => add #27y# to both sides:
#y^3-9y^2+27y-27=0# => factor by grouping:
#y^3-27-9y^2+27y=0# => difference of two cubes:
#(y-3)(y^2+3y+9)-9y(y-3)=0# =>factor #y-3#:
#(y - 3)(y^2+3y+9-9y)=0# => simplify:
#(y-3)(y^2-6y+9)=0#
#(y-3)(y-3)^2=0#
#(y-3)^3=0#
#y-3=0#
#y=3#

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