What is the sum of all the values of x?

1. Let x^3=(a+1/a)(b+1/b)(c+1/c)=abc+1/(abc)+a+1/a+b+1/b+c+1/c
2. x^3=abc+1/(abc)+3x
3. Replace abc with N
4. x^3-0*x^2-3x-N-1/N=0
5. The sum of the roots of a cubic, by Vieta’s formula=-b/a=0/1=0

$x=a+ \frac{1}{b} \rightarrow b= \frac {1}{x-a} \newline x=b+ \frac{1}{c} \rightarrow c= \frac{1}{x-b} = \frac{1}{x-\frac{1}{x-a}} \newline x=c+\frac{1}{a} \rightarrow a=\frac{1}{x-c} = \LARGE{\frac{1}{x-\frac{1}{x-\frac{1}{x-a}}}} \newline =\LARGE{\frac{1}{x-\frac{x-a}{x^2-ax-1}}} \newline =\LARGE{\frac{x^2-ax-1}{x^3-ax^2-2x+a}} \newline$ $a \times (x^3-ax^2-2x+a) = x^2-ax-1 \newline ax^3-a^2x^2-2ax+a^2=x^2-ax-1 \newline ax^3-a^2x^2- 2ax+a^2-x^2+ ax+1=0 \newline ax^3-a^2x^2- ax+a^2-x^2+1=0 \newline ax^3-a^2x^2-x^2-ax+a^2+1=0 \newline ax \times x^2 - a^2 \times x^2 - 1 \times x^2- ax + a^2 + 1 =0 \newline x^2(ax-a^2-1)-(ax-a^2-1)=0 \newline (x^2-1)(ax-a^2-1)=0 \newline x^2-1=0$ or $ax-a^2-1=0. \newline$ If $ax-a^2-1=0,$ then $ax-a^2=1 \newline a(x-a)=1 \newline x-a=\frac{1}{a} \newline x=a+\frac{1}{a} \rightarrow a=b$ . Does not match the meaning of the title. So $ax-a^2-1 \neq 0 \rightarrow x^2-1=0 \newline x^2=1 \newline x=1$ or $x=-1$ . The sum of all the values of x is $1+(-1)=\boxed{0}$

For any solution $x_1=x(a , b, c)$ there is also a solution $x_2= x(-a,-b,-c)=-x(a,b,c)=-x_1$ . The solutions come in pairs of opposite sign. If there is a finite number of solutions, $\sum x =0$ .

Note that $a-b = b^{-1}c^{-1}(b-c) \hspace{1cm} b-c = a^{-1}c^{-1}(c-a) \hspace{1cm} c-a = a^{-1}b^{-1}(a-b)$ so that $a-b = (abc)^{-2}(a-b)$ and hence $abc = \pm1$ .

• If $abc=1$ then $a-b = a(b-c)$ and $b-c = b(c-a)$ , so that $0 = (a-b)+(b-c)+(c-a) = (ab + b + 1)(c-a)$ , and hence $ab + b + 1 = 0$ , so that $x = a + b^{-1} = -1$ . Note that $a=1$ , $b=-\tfrac12$ , $c=-2$ give an example of this case.

• If $abc=-1$ then $a-b=-a(b-c)$ and $b-c=-b(c-a)$ , so that $0 = (a-b)+(b-c)+(c-a) = (ab - b + 1)(c-a)$ , and hence $ab - b + 1 = 0$ , so that $x = a + b^{-1} = 1$ . Note that $a=2$ , $b=-1$ , $c=\tfrac12$ give an example of this case.

Thus the answer is $-1 + 1 = \boxed{0}$ .