# How do you solve 1/x+1/x^2+1/x^3=87 ?

Nov 6, 2017

$\frac{1}{x} + \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3 = 87$

Put everything with the same denominator:

${x}^{2} / {x}^{3} + \frac{x}{x} ^ 3 + \frac{1}{x} ^ 3 = 87 {x}^{3} / {x}^{3}$

Simplify, considering $x \ne 0$:

${x}^{2} + x + 1 = 87 {x}^{3}$

Arranging the factors:

$87 {x}^{3} - {x}^{2} - x - 1 = 0$

Considering that this an equation with only integers the solutions are contained on the division of the divisors of -1 (the independent variable) by the divisors of the factor of highest degree 87.

$87 = 3 \cdot 29$ so will we have $\pm 1$, $\pm 3$, $\pm 29$ and $\pm 87$

And the factors to experiment will be

$\pm \frac{1}{1}$, $\pm \frac{1}{3}$, $\pm \frac{1}{29}$ and $\pm \frac{1}{87}$

The easier equivalent form of the equation is this:

$87 {x}^{3} - {x}^{2} - x - 1 = 0$

So, let's try:

## Positive numbers:

$87 - 1 - 1 - 1 = 84$

$87 \cdot 3 - {3}^{2} - 3 - 1 = 248$

$87 \cdot 29 - {29}^{2} - 29 - 1 = 2232$

$87 \cdot 87 - {87}^{2} - 87 - 1 = - 88$

## Negative numbers:

$- 87 - 1 + 1 - 1 = - 88$

$- 87 \cdot 3 - {3}^{2} + 3 - 1 = - 268$

$- 87 \cdot 29 - {29}^{2} + 29 - 1 = - 3336$

$- 87 \cdot 87 - {87}^{2} + 87 - 1 = - 15224$

Nov 6, 2017

Real root:

$x = \frac{1}{261} \left(1 + \sqrt{102574 + 1566 \sqrt{4283}} + \sqrt{102574 - 1566 \sqrt{4283}}\right)$

and related complex roots...

#### Explanation:

Given:

$\frac{1}{x} + \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3 = 87$

Multiply both sides by ${x}^{3}$ to get:

${x}^{2} + x + 1 = 87 {x}^{3}$

Subtract ${x}^{2} + x + 1$ from both sides to get:

$87 {x}^{3} - {x}^{2} - x - 1 = 0$

Multiply by $204363 = {87}^{2} \cdot {3}^{3}$ to avoid fractions and find:

$0 = 204363 \cdot \left(87 {x}^{3} - {x}^{2} - x - 1\right)$

$\textcolor{w h i t e}{0} = 17779581 {x}^{3} - 204363 {x}^{2} - 204363 x - 204363$

$\textcolor{w h i t e}{0} = {\left(261 x\right)}^{3} - 3 {\left(261 x\right)}^{2} + 3 \left(261 x\right) - 1 - 786 \left(261 x\right) + 786 - 205148$

$\textcolor{w h i t e}{0} = {\left(261 x - 1\right)}^{3} - 786 \left(261 x - 1\right) - 205148$

$\textcolor{w h i t e}{0} = {t}^{3} - 786 t - 205148$

where $t = 261 x - 1$

Using Cardano's method, let $t = u + v$ to get:

${u}^{3} + {v}^{3} + 3 \left(u v - 262\right) \left(u + v\right) - 205148 = 0$

To eliminate the term in $\left(u + v\right)$ add the constraint:

$u v - 262 = 0 \text{ }$ i.e. $v = \frac{262}{u}$

Then our equation becomes:

${u}^{3} + \frac{17984728}{u} ^ 3 - 205148 = 0$

Multiply through by ${u}^{3}$ and rearrange slightly to get:

${\left({u}^{3}\right)}^{2} - 205148 \left({u}^{3}\right) + 17984728 = 0$

Using the quadratic formula, we find:

${u}^{3} = \frac{205148 \pm \sqrt{{205148}^{2} - 4 \left(1\right) \left(17984728\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{205148 \pm \sqrt{42085701904 - 71938912}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{205148 \pm \sqrt{42013762992}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{205148 \pm 3132 \sqrt{4283}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = 102574 \pm 1566 \sqrt{4283}$

Now since the derivation was symmetric and these roots are real, we can use one of these roots as ${u}^{3}$ and the other as ${v}^{3}$ to find real root of our cubic in $t$:

${t}_{1} = \sqrt{102574 + 1566 \sqrt{4283}} + \sqrt{102574 - 1566 \sqrt{4283}}$

and related complex roots:

${t}_{2} = \omega \sqrt{102574 + 1566 \sqrt{4283}} + {\omega}^{2} \sqrt{102574 - 1566 \sqrt{4283}}$

${t}_{3} = {\omega}^{2} \sqrt{102574 + 1566 \sqrt{4283}} + \omega \sqrt{102574 - 1566 \sqrt{4283}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive complex cube root of $1$.

Then $x = \frac{1}{261} \left(1 + t\right)$

So the roots of our original equation are:

${x}_{1} = \frac{1}{261} \left(1 + \sqrt{102574 + 1566 \sqrt{4283}} + \sqrt{102574 - 1566 \sqrt{4283}}\right)$

${x}_{2} = \frac{1}{261} \left(1 + \omega \sqrt{102574 + 1566 \sqrt{4283}} + {\omega}^{2} \sqrt{102574 - 1566 \sqrt{4283}}\right)$

${x}_{3} = \frac{1}{261} \left(1 + {\omega}^{2} \sqrt{102574 + 1566 \sqrt{4283}} + \omega \sqrt{102574 - 1566 \sqrt{4283}}\right)$