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 $omega = -1/2+sqrt(3)/2i$

How do you factor 64y^3 + 2764y^3 + 27?