How do you factor this polynomial completely? #x^3 + 3x^2y + 3xy^2 + y^3 - #64

Question

#x^3 + 3x^2y + 3xy^2 + y^3 - #64

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Answer 1 · 2018-07-08T07:16:54

Please see the explanation below

We need

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

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

Therefore,

#x^3+3x^2y+3xy^2+y^3-64#

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

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

#=(x+y-4)((x+y)(x+y+4)+16)#

Hope that this will help!!!