We have,
$$
\begin{aligned}
\quad(x-i y)^{\frac{1}{3}} & =a+i b \\
\Rightarrow \quad x-i y & =(a+i b)^3 \\
\Rightarrow \quad x-i y & =a^3+3 a^2 b i+3 a b^2 i^2+i^3 b^3 \\
\Rightarrow \quad x-i y & =\left(a^3-3 a b^2\right)+\left(3 a^2 b-b^3\right) i \\
\therefore \quad x & =a^3-3 a b^2 \text { and } y=b^3-3 a^2 b \\
\text { Now, } \frac{a x-b y}{a-b} & =\frac{a\left(a^3-3 a b^2\right)-b\left(b^3-3 a^2 b\right)}{a-b} \\
= & \frac{a^4-3 a^2 b^2-b^4+3 a^2 b^2}{a-b} \\
& =\frac{\left(a^4-b^4\right)}{(a-b)}=\frac{(a-b)(a+b)\left(a^2+b^2\right)}{(a-b)} \\
& =(a+b)\left(a^2+b^2\right)=a^3+a^2 b+a b^2+b^3
\end{aligned}
$$