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

#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#

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)))#