Question

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

Answer 1

`1/x+1/x^2+1/x^3=87`

Put everything with the same denominator:

`x^2/x^3+x/x^3+1/x^3=87x^3/x^3`

Simplify, considering `x!=0`:

`x^2+x+1=87x^3`

Arranging the factors:

`87x^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*29` so will we have `+-1`, `+-3`, `+-29` and `+-87`

And the factors to experiment will be

`+-1/1`, `+-1/3`, `+-1/29` and `+-1/87`

The easier equivalent form of the equation is this:

`87x^3-x^2-x-1=0`

So, let's try:

Positive numbers:

`87-1-1-1=84`

`87*3-3^2-3-1=248`

`87*29-29^2-29-1=2232`

`87*87-87^2-87-1=-88`

Negative numbers:

`-87-1+1-1=-88`

`-87*3-3^2+3-1=-268`

`-87*29-29^2+29-1=-3336`

`-87*87-87^2+87-1=-15224`

Answer 2

Real root:

`x = 1/261(1+root(3)(102574+1566sqrt(4283))+root(3)(102574-1566sqrt(4283)))`

and related complex roots...

Explanation:

Given:

`1/x+1/x^2+1/x^3=87`

Multiply both sides by `x^3` to get:

`x^2+x+1=87x^3`

Subtract `x^2+x+1` from both sides to get:

`87x^3-x^2-x-1=0`

Multiply by `204363 = 87^2*3^3` to avoid fractions and find:

`0 = 204363*(87x^3-x^2-x-1)`

`color(white)(0) = 17779581x^3-204363x^2-204363x-204363`

`color(white)(0) = (261x)^3-3(261x)^2+3(261x)-1-786(261x)+786-205148`

`color(white)(0) = (261x-1)^3-786(261x-1)-205148`

`color(white)(0) = t^3-786t-205148`

where `t = 261x-1`

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

`u^3+v^3+3(uv-262)(u+v)-205148 = 0`

To eliminate the term in `(u+v)` add the constraint:

`uv-262 = 0" "` i.e. `v=262/u`

Then our equation becomes:

`u^3+17984728/u^3-205148 = 0`

Multiply through by `u^3` and rearrange slightly to get:

`(u^3)^2-205148(u^3)+17984728=0`

Using the quadratic formula, we find:

`u^3 = (205148+-sqrt(205148^2-4(1)(17984728)))/(2*1)`

`color(white)(u^3) = (205148+-sqrt(42085701904-71938912))/2`

`color(white)(u^3) = (205148+-sqrt(42013762992))/2`

`color(white)(u^3) = (205148+-3132sqrt(4283))/2`

`color(white)(u^3) = 102574+-1566sqrt(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 = root(3)(102574+1566sqrt(4283))+root(3)(102574-1566sqrt(4283))`

and related complex roots:

`t_2 = omega root(3)(102574+1566sqrt(4283))+omega^2 root(3)(102574-1566sqrt(4283))`

`t_3 = omega^2 root(3)(102574+1566sqrt(4283))+omega root(3)(102574-1566sqrt(4283))`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`.

Then `x = 1/261(1+t)`

So the roots of our original equation are:

`x_1 = 1/261(1+root(3)(102574+1566sqrt(4283))+root(3)(102574-1566sqrt(4283)))`

`x_2 = 1/261(1+omega root(3)(102574+1566sqrt(4283))+omega^2 root(3)(102574-1566sqrt(4283)))`

`x_3 = 1/261(1+omega^2 root(3)(102574+1566sqrt(4283))+omega root(3)(102574-1566sqrt(4283)))`