How do you factor #64y^3 + 27#?

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Answer 1 · 2015-05-22T23:40:03

Using #a^3+b^3 = (a+b)(a^2-ab+b^2)# we find:

#64y^3+27 = ((4y)^3+3^3)#

#= (4y+3)((4y)^2 - (4y)*3 + 3^2)#

#= (4y+3)(16y^2-12y+9)#

The factor #(16y^2-12y+9)# has no linear factors with real coefficients:

The discriminant

#Delta(16y^2-12y+9) = (-12)^2-(4xx16xx9)#
#=144-576=-432 < 0#

...so #16y^2-12y+9 = 0# has no real roots.

If you are curious, the other (complex) linear factors are actually

#(4y+3omega)(4y+3omega^2)#, where #omega = -1/2+sqrt(3)/2i#