# How do you factor x^3+27?

Nov 3, 2016

Because factoring x^3+27 is the same as finding where the graph passes through the x axis, we can just set the equation equal to zero and solve. $f \left(x\right) = \left(x + 3\right) \left({x}^{2} - 3 x + 9\right)$

#### Explanation:

Let f(x) = x³ + 27
0 = x³ + 27
x³= -27
x = -3
This means that x = -3 is the only zero of the graph of f(x). Since we know $\left(x + 3\right)$ is one factor of $f \left(x\right)$, to find the 2nd factor,
$\frac{{x}^{3} + 27}{x + 3}$
We get the second factor to be ${x}^{2} - 3 x + 9$.

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

Nov 3, 2016

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

#### Explanation:

The sum of cubes identity can be written:

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

Note that both ${x}^{3}$ and $27 = {3}^{3}$ are perfect cubes, so we can use the sum of cubes identity with $a = x$ and $b = 3$ as follows:

${x}^{3} + 27 = {x}^{3} + {3}^{3}$

$\textcolor{w h i t e}{{x}^{3} + 27} = \left(x + 3\right) \left({x}^{2} - 3 x + {3}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + 27} = \left(x + 3\right) \left({x}^{2} - 3 x + 9\right)$

The remaining quadratic has no simpler linear factors with Real coefficients.