# How do you factor 64x^3 + 343?

Oct 31, 2017

$64 {x}^{3} + 343 = \left(4 x + 7\right) \left(16 {x}^{2} - 28 x + 49\right)$

#### Explanation:

Note that $64 {x}^{3} = {\left(4 x\right)}^{3}$ and $343 = {7}^{3}$ are both perfect cubes and:

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

So putting $a = 4 x$ and $b = 7$ we find:

$64 {x}^{3} + 343 = {\left(4 x\right)}^{3} + {7}^{3}$

$\textcolor{w h i t e}{64 {x}^{3} + 343} = \left(4 x + 7\right) \left({\left(4 x\right)}^{2} - \left(4 x\right) \left(7\right) + {7}^{2}\right)$

$\textcolor{w h i t e}{64 {x}^{3} + 343} = \left(4 x + 7\right) \left(16 {x}^{2} - 28 x + 49\right)$

The remaining quadratic factor has no linear factors with real coefficients. It can be factored using the primitive complex cube root of $1$:

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

Then:

$16 {x}^{2} - 28 x + 49 = \left(4 x + 7 \omega\right) \left(4 x + 7 {\omega}^{2}\right)$