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

##### 2 Answers
May 23, 2015

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

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

May 23, 2015

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

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

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

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

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

To check this, evaluate its discriminant:

$\Delta \left(16 {x}^{2} - 12 x + 9\right) = {\left(- 12\right)}^{2} - \left(4 \times 16 \times 9\right)$

$= 144 - 576 = - 432$

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

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

$\left(4 x + 3 \omega\right)$ and $\left(4 x + 3 {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$