# How do you factor y= 3x^3 + 81  ?

Mar 6, 2018

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

#### Explanation:

$y = 3 {x}^{3} + 81$

$\implies y = 3 \left({x}^{3} + 27\right)$

$\implies y = 3 \left({x}^{3} + {3}^{3}\right)$

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

So , $y = 3 \left({x}^{3} + {3}^{3}\right)$

$\implies y = 3 \left(x + 3\right) \left({x}^{2} + {3}^{2} - x \times 3\right)$

$\implies y = 3 \left(x + 3\right) \left({x}^{2} + 9 - 3 x\right)$