Let $u=ax+by+a\sqrt[3]{b}=0,v=bx-ay+b\sqrt[3]{a}=0$ be two straight lines. The equation of the bisectors of the angle formed by $L_1\equiv \left(\tan {\theta }_1\right)u-\left(\tan {\theta }_2\right)v=0$ and $L_2\equiv \left(\tan {\theta }_1\right)u+\left(\tan {\theta }_2\right)v=0\text{ for }{\theta }_1,{\theta }_2\in \left(0,\frac{\pi }{2}\right)$ is 

Note that lines $u = 0, v = 0$ are perpendicular. 

Make the co-ordinate axes coincide with $u=0$

$v = 0.$

Now the lines $L_1\equiv 0,L_2\equiv 0$ are equally

inclined with UV-axes

$\therefore u=0,v=0$ are bisectors.