How do you factor completely #64 - 27y^3 #?

1 Answer
George C.
Aug 5, 2016

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

Explanation:

The difference of cubes identity can be written:

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

Note that #64=4^3# and #27y^3 = (3y)^3# are both perfect cubes, so we find:

#64-27y^3#

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

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

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

The remaining quadratic has negative discriminant. It cannot be factored further with Real coefficients. If we allow Complex coefficients then we can factor it completely:

#= (4-y)(4-omega y)(4-omega^2 y)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

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