$${ \left(1a/2+2b/3 \right) }^{ 3 }$$
$\frac{\left(3a+4b\right)^{3}}{216}$
$\frac{ba^{2}}{2}+\frac{2ab^{2}}{3}+\frac{8b^{3}}{27}+\frac{a^{3}}{8}$
