Let $a=\frac{1}{2}, b=\frac{1}{3}, c=-\frac{5}{6}$

$\therefore \quad a+b+c=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$

$=\frac{3+2-5}{6}=\frac{0}{6}=0$

$\Rightarrow \quad a^{3}+b^{3}+c^{3}=3 a b c$

$\therefore \quad\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}=\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(-\frac{5}{6}\right)^{3}$

$=3 \times \frac{1}{2} \times \frac{1}{3}\left(-\frac{5}{6}\right)=-\frac{5}{12}$