# How do you factor 125x^3+8g^3?

Feb 13, 2017

$125 {x}^{3} + 8 {g}^{3} = \left(5 x + 2 g\right) \left(25 {x}^{2} - 10 x g + 4 {g}^{2}\right)$

#### Explanation:

As $125 {x}^{3} + 8 {g}^{3} = {\left(5 x\right)}^{3} + {\left(2 g\right)}^{3}$, it can be factorized using identity ${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

and $125 {x}^{3} + 8 {g}^{3} = {\left(5 x\right)}^{3} + {\left(2 g\right)}^{3}$

= $\left(5 x + 2 g\right) \left({\left(5 x\right)}^{2} - \left(5 x\right) \times \left(2 g\right) + {\left(2 g\right)}^{2}\right)$

= $\left(5 x + 2 g\right) \left(25 {x}^{2} - 10 x g + 4 {g}^{2}\right)$

For a proof of the identity see below.

${x}^{3} + {y}^{3}$

= ${x}^{3} + {x}^{2} y - {x}^{2} y - x {y}^{2} + x {y}^{2} + {y}^{3}$

= ${x}^{2} \left(x + y\right) - x y \left(x + y\right) + {y}^{2} \left(x + y\right)$

= $\left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$