# How do you factor  27g^3 + 343?

Mar 11, 2016

$27 {g}^{3} + 343 = \left(3 g + 7\right) \left(9 {g}^{2} - 21 g + 49\right)$

#### Explanation:

We need to write the expression in a form that remembers the remarkable identities.

In this case we could try to see if $343$ is a perfect cube.

$343 = {7}^{3}$

$\therefore 27 {g}^{3} + 343 = {\left(3 g\right)}^{3} + {7}^{3}$

Now it's a sum of cubes then we could use the following identity:

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

$\therefore {\left(3 g\right)}^{3} + {7}^{3} = \left(3 g + 7\right) \left({\left(3 g\right)}^{2} - 3 g \cdot 7 + {7}^{2}\right) =$
$= \left(3 g + 7\right) \left(9 {g}^{2} - 21 g + 49\right)$