# How do you factor 27m^3+1?

Jul 3, 2016

$\left(3 m + 1\right) \left(9 {m}^{2} - 3 m + 1\right)$

#### Explanation:

This is a $\textcolor{b l u e}{\text{sum of cubes}}$ and is factorised in general as follows.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$27 {m}^{3} = {\left(3 m\right)}^{3} \text{ and " 1=(1)^3rArra=3m" and } b = 1$

Substituting these values gives required factors.

$\Rightarrow 27 {m}^{3} + 1 = \left(3 m + 1\right) \left(9 {m}^{2} - 3 m + 1\right)$

Jul 3, 2016

Using the formula for the sum of cubes.

#### Explanation:

We can use the formula for the sum of cubes

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

here our cubes are

${\left(3 m\right)}^{3} + {1}^{3} = \left(3 m + 1\right) \left({\left(3 m\right)}^{2} - 3 m + 1\right)$

$= \left(3 m + 1\right) \left(9 {m}^{2} - 3 m + 1\right)$.

The roots of $9 {m}^{2} - 3 m + 1$ are

$m = \frac{3 \setminus \pm \sqrt{9 - 36}}{18} = \frac{3 \setminus \pm \sqrt{- 27}}{18}$

$= \frac{3 \setminus \pm i 3 \sqrt{3}}{18} = \frac{1 \setminus \pm i \sqrt{3}}{6}$

then are both complex. This means that the quadratic part cannot be factorized more with real numbers. We can write the irreducible factorization in complex numbers as

$\left(3 m + 1\right) \left(m - \frac{1 + i \sqrt{3}}{6}\right) \left(m - \frac{1 - i \sqrt{3}}{6}\right)$.