Hexotic Equation - 1

$\begin{aligned} 6x^6 - 25x^5 + 31x^4 - 31x^2 + 25x - 6 & = 0 \\ 6(x^6 - 1) - 25x(x^4-1) + 31x^2(x^2-1) & = 0 \\ 6(x^2-1)(x^4+x^2+1) - 25x(x^2-1)(x^2+1) +31x^2(x^2-1) & = 0 \\ (x^2-1)(6x^4-25x^3+37x^2-25x+6) & = 0 \\ (x^2-1)(6x^4-12x^3-13x^3+26x^2+11x^2-22x-3x+6) & = 0 \\ (x^2-1)(6x^3(x-2)-13x^2(x-2)+11x(x-2)-3(x-2)) & = 0 \\ (x^2-1)(x-2)(6x^3-13x^2+11x-3) & = 0 \\ (x^2-1)(x-2)(6x^3-3x^2-10x^2+5x+6x-3) & = 0 \\ (x^2-1)(x-2)(3x^2(2x-1)-5x(2x-1)+3(2x-1)) & = 0 \\ (x^2-1)(x-2)(2x-1)(3x^2-5x + 3) & = 0 \end{aligned}$

We note that $3x^2-5x + 3 = 0$ has no real root, therefore the sum of integral real roots is $1-1+2 = \boxed{2}$

Obviously $x=0$ is not a solution of the given equation. So dividing both sides of the equation by $x^{3}$ we obtain the equivalent equation $6x^3-25x^2+31x-\frac{31}{x}+\frac{25}{x^2}-\frac{6}{x^{3}}=0.$ Rearranging terms we get $6(x^3-\frac{1}{x^{3}})-25(x^2-\frac{1}{x^2})+31(x-\frac{1}{x})=0.$ Factoring $x-\frac{1}{x}$ out we get $(x-\frac{1}{x})(6(x^2+\frac{1}{x^{2}}+1)-25(x+\frac{1}{x})+31)=0.$ The latter equation can be rewritten as $(x-\frac{1}{x})(6(x+\frac{1}{x})^{2}-25(x+\frac{1}{x})+25)=0.$ Therefore we get that $x-\frac{1}{x}=0,$ $x+\frac{1}{x}=\frac{5}{3},$ and $x+\frac{1}{x}=\frac{5}{2}.$ Solving these three equations we get that the solutions are $1, -1, 2, \frac{1}{2}, \frac{5+i\sqrt{11}}{6},$ and $\frac{5-i\sqrt{11}}{6}.$ Therefore the sum of the integer roots is $1+(-1)+2=2.$

Using trial and error (use the factors of 6) , we get x=1,or -1,
6x^6 – 25x^5 + 31x^4 – 31x^2 + 25x – 6 = (x^2 – 1)(6x^4 - 25x^3 + 37x^2 - 25x + 6) =0 Repeat again, then x=2, So (6x^4 - 25x^3 + 37x^2 - 25x + 6) = (x – 2)(6x^3 - 13x^2 + 11x – 3) = 0 Also x=1/2,is the solution of (6x^3 - 13x^2 + 11x – 3) = (2x – 1)(3x^2 - 5x +3) = 0.So the sum of integral roots = 1 - 1 + 2 = 2