How do you factor $64 x^{3} + 27$ $64x^3+27$ ?

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Answer 1 · 2015-05-23T22:56:25

Use the identity: $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3 + b^3 = (a +b)(a^2 - ab + b^2)$

f(x) = (4x + 3)(16x^2 - 12x + 9) =

Answer 2 · 2015-05-23T22:58:29

Using the identity $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$ we find:

$64 x^{3} + 27 = {(4 x)}^{3} + 3^{3}$ $64x^3+27 = (4x)^3 + 3^3$

$= (4 x + 3) ({(4 x)}^{2} - (4 x) \cdot 3 + 3^{2})$ $= (4x+3)((4x)^2-(4x)*3+3^2)$

$= (4 x + 3) (16 x^{2} - 12 x + 9)$ $= (4x+3)(16x^2-12x+9)$

$16 x^{2} - 12 x + 9$ $16x^2-12x+9$ has no simpler factors with real coefficients.

To check this, evaluate its discriminant:

$Delta(16x^2-12x+9) = (-12)^2-(4xx16xx9)$

$= 144-576 = -432$

Since $Delta < 0$ the quadratic equation $16x^2-12x+9 = 0$ has no real roots. It has two distinct complex roots.

In case you are curious, the complex factors of $16x^2-12x+9$ are:

$(4x+3omega)$ and $(4x+3omega^2)$

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

How do you factor 64x^3+2764x3+2764x^3+27?