# How do you factor  x^3 - 1?

Dec 15, 2017

#### Explanation:

I want to expand upon an idea expressed in the prior answer

The idea of:

$\frac{{x}^{n} - 1}{x - 1} = {\sum}_{r = 1}^{n} {x}^{n - r}$

or not in sigma notation:

$\frac{{x}^{n} - 1}{x - 1} = {x}^{n - 1} + {x}^{n - 2} + \ldots + x + 1$

We can prove this via induction:

Basis case :

$\implies n = 1$

$L H S : \frac{{x}^{1} - 1}{x - 1} = 1$

$R H S : {x}^{1 - 1} = {x}^{0} = 1$

Hence basis case holds

Induction:

Assume $n = k$ holds:

$\frac{{x}^{k} - 1}{x - 1} = {\sum}_{r = 1}^{k} {x}^{k - r}$

$n = k + 1$:

${\sum}_{r = 1}^{k + 1} {x}^{k + 1 - r} = \left({\sum}_{r = 1}^{k} {x}^{k + 1 - r}\right) + 1$

$= x \cdot \left({\sum}_{r = 1}^{k} {x}^{k - r}\right) + 1$

$= x \cdot \left(\frac{{x}^{k} - 1}{x - 1}\right) + 1$

$= \frac{{x}^{k + 1} - x}{x - 1} + 1$

$= \frac{{x}^{k + 1} - x}{x - 1} + \frac{x - 1}{x - 1}$

$= \frac{{x}^{k + 1} - 1}{x - 1}$

Hence this is also what we yield when plugging directly into formula:

Hence holds for all $k \in {\mathbb{Z}}^{+}$ and all $k + 1 \in {\mathbb{Z}}^{+}$ so holds for all $n \in {\mathbb{Z}}^{+}$

$\implies$ Proven by mathematical induction

I thought this was a nice idea to consider!

Dec 30, 2017

$\left(x - 1\right) \left({x}^{2} + x + 1\right)$

#### Explanation:

${x}^{3} - 1 \text{ is a "color(blue)"difference of cubes}$

•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)

$\text{here "a=x" and } b = 1$

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