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

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Answer 1 · 2018-05-03T15:22:58

#(4x+y)(16x^2-4xy+y^2)#

This is an example of factoring perfect cubes.

#64x^3+y^3#

#64# is a perfect cube #4*4*4#
#x^3# is a perfect cube #x*x*x#
#y^3# is a perfect cube #y*y*y#

The factor pattern for factoring perfect cubes is always

#(a+b)(a^2-ab+b^2)#
or
#(a-b)(a^2+ab+b^2)#

The signs follow the acronym SOAP

First sign is the SAME as the expression
Second sign is the OPPOSITE of the expression
Third sign is ALWAYS POSITIVE

For this problem
#a =4x# the root of #64x^3#
#b=y# the root of #y^3#

So plugging in we get

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

#(4x+y)(16^2 - 4xy +y^3)#

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