# Question #a407d

Jun 1, 2016

${x}^{2014} + {x}^{2013} = - {x}^{2015} = - {x}^{2012}$

#### Explanation:

Take the expression ${x}^{2014} + {x}^{2013}$ and factor ${x}^{2012}$ from each term.

${x}^{2014} + {x}^{2013} = {x}^{2012} \left({x}^{2} + x\right)$

Note that from the equation ${x}^{2} + x + 1 = 0$ we can state that ${x}^{2} + x = - 1$. Thus,

${x}^{2012} \left({x}^{2} + x\right) = {x}^{2012} \left(- 1\right) = - {x}^{2012}$

Alternatively, start by factoring ${x}^{2013}$ from both terms.

${x}^{2014} + {x}^{2013} = {x}^{2013} \left(x + 1\right)$

From ${x}^{2} + x + 1 = 0$ we see that $x + 1 = - {x}^{2}$, so:

${x}^{2013} \left(x + 1\right) = {x}^{2013} \left(- {x}^{2}\right) = - {x}^{2015}$

Another method:

${x}^{2014} + {x}^{2013} = {x}^{2014} + {x}^{2013} + {x}^{2012} - {x}^{2012}$

Factor ${x}^{2012}$ from the first three terms:

${x}^{2014} + {x}^{2013} + {x}^{2012} - {x}^{2012} = {x}^{2012} \left({x}^{2} + x + 1\right) - {x}^{2012}$

Since ${x}^{2} + x + 1 = 0$, this equals

${x}^{2012} \left({x}^{2} + x + 1\right) - {x}^{2012} = 0 - {x}^{2012} = - {x}^{2012}$

Similar to this, we can add and subtract ${x}^{2015}$:

${x}^{2014} + {x}^{2013} = {x}^{2015} + {x}^{2014} + {x}^{2013} - {x}^{2015}$

Factor ${x}^{2013}$ from the first three terms:

${x}^{2015} + {x}^{2014} + {x}^{2013} - {x}^{2015} = {x}^{2013} \left({x}^{2} + x + 1\right) - {x}^{2015}$

Using the same logic as before,

${x}^{2013} \left({x}^{2} + x + 1\right) - {x}^{2015} = - {x}^{2015}$

Jun 1, 2016

${x}^{2014} + {x}^{2013} = x + 1 = \frac{1 \pm i \sqrt{3}}{2}$

#### Explanation:

${x}^{2} + x + 1$ is the quadratic factor of ${x}^{3} - 1$.

So ${x}^{2} + x + 1 = 0$ implies that ${x}^{3} - 1 = 0$.

(If necessary for clarity, multiply both sides by $x - 1$.)

So we see that ${x}^{3} = 1$.

$\frac{2013}{3} = 671$, so

${x}^{2013} = {\left({x}^{3}\right)}^{671} = {1}^{671} = 1$

Method 1

${x}^{2014} = x \cdot {x}^{2013} = x \cdot 1 = x$

${x}^{2014} + {x}^{2013} = x + 1$

Method 2

${x}^{2014} + {x}^{2013} = {x}^{2013} \left(x + 1\right) = 1 \cdot \left(x + 1\right) = x + 1$

If you want a numerical answer , solve ${x}^{2} + x + 1 = 0$ for $x = \frac{- 1 \pm i \sqrt{3}}{2}$

We conclude with

$x + 1 = \frac{1 \pm i \sqrt{3}}{2}$.