If (x-iy)^{frac{1}{3}} = a+ib,then frac{ax-by | vedclass

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(B) Given $(x-iy)^{\frac{1}{3}} = a+ib$. Cubing both sides,we get $x-iy = (a+ib)^3$. Expanding the right side,$x-iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)

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$a^3+a^2b+ab^2+b^3$

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(B) Given $(x-iy)^{\frac{1}{3}} = a+ib$. Cubing both sides,we get $x-iy = (a+ib)^3$. Expanding the right side,$x-iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$. Since $i^2 = -1$ and $i^3 = -i$,we have $x-iy = a^3 + 3a^2bi - 3ab^2 - ib^3$. Grouping real and imaginary parts,$x-iy = (a^3-3ab^2) - i(b^3-3a^2b)$. Comparing real and imaginary parts,$x = a^3-3ab^2$ and $y = b^3-3a^2b$. Now,substitute these into the expression $\frac{ax-by}{a-b}$: $\frac{a(a^3-3ab^2) - b(b^3-3a^2b)}{a-b} = \frac{a^4-3a^2b^2 - b^4+3a^2b^2}{a-b}$. $= \frac{a^4-b^4}{a-b} = \frac{(a-b)(a+b)(a^2+b^2)}{a-b}$. $= (a+b)(a^2+b^2) = a^3+ab^2+a^2b+b^3 = a^3+a^2b+ab^2+b^3$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

If $(x-iy)^{\frac{1}{3}} = a+ib$,then $\frac{ax-by}{a-b} = $

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