Let $\alpha, \beta$ be the roots of the quadratic equation $x^{2}+\sqrt{6}x+3=0$. Then $\dfrac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to: