Given equation,
$\begin{aligned}
& 6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0 \\
& 6\left(x^6-1\right)-25 x\left(x^4-1\right)+31\left(x^4-x^2\right)=0 \\
& \left(x^2-1\right)\left[6 x^4+6 x^2+6-25 x^3-25 x+31 x^2\right]=0 \\
& 6 x^4+6-25\left(x^3+x\right)+37 x^2=0 \\
& 6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)+37=0 \\
& 6\left(x+\frac{1}{x}\right)^2-12-25\left(x+\frac{1}{x}\right)+37=0 \\
& 6\left(x+\frac{1}{x}\right)^2-25\left(x+\frac{1}{x}\right)+25=0 \\
& \therefore 6 y^2-25 y+25=0 \quad\left[\because x+\frac{1}{x}=y\right] x \\
& (6 y-15)(6 y-10)=0 \\
& y=\frac{5}{2}, \frac{5}{3} \Rightarrow x+\frac{1}{x}=\frac{5}{2} \Rightarrow x=2, \frac{1}{2} \\
& x+\frac{1}{x}=\frac{5}{3} \text { has no real roots } \\
& \therefore \quad \alpha+\beta=\frac{5}{3} \\
&
\end{aligned}$