# How do you factor 27+8x^3?

Feb 5, 2016

$27 + 8 {x}^{3} = \left(3 + 2 x\right) \left(9 - 6 x + 4 {x}^{2}\right) = \left(2 x + 3\right) \left(4 {x}^{2} - 6 x + 9\right)$

#### Explanation:

The sum of two cubes can be factored as:

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

For the problem at hand, $27 + 8 {x}^{3} = {3}^{3} + {\left(2 x\right)}^{3}$ so that $a = 3$ and $b = 2 x$. Therefore

$27 + 8 {x}^{3} = \left(3 + 2 x\right) \left(9 - 6 x + 4 {x}^{2}\right) = \left(2 x + 3\right) \left(4 {x}^{2} - 6 x + 9\right)$

It is 27+8x^3= (3+2x)*(9-6x+4x^2)
27+8x^3=(3^3)+(2x)^3=(3+2x)*(3^2-6x+4x^2)= (3+2x)*(9-6x+4x^2)