jmc

Clearly, $x=0$ is not a root. We can divide the equation by $x^3,$ to get $$x^3+x-115+\frac{1}{x}+\frac{1}{x^3}=0.$$Let $y=x+\frac{1}{x}.$ Then $$y^3=x^3+3x+\frac{3}{x}+\frac{1}{x^3},$$so $$x^3+\frac{1}{x^3}=y^3-3\left(x+\frac{1}{x}\right)=y^3-3y.$$Thus, our equation becomes $$y^3-3y+y-115=0,$$or $y^3-2y-115=0.$ This equation factors as $(y-5)(y^2+5y+23)=0.$ The quadratic factor has no real roots, so $y=5.$ Then $$x+\frac{1}{x}=5,$$or $x^2-5x+1=0.$ This quadratic does have real roots, and by Vieta's formulas, their sum is $\boxed{5}.$