If (x+iy)^{1/3} = a+ib where x, y, a, b in R | vedclass

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(A) Given $(x+iy)^{1/3} = a+ib$.Cubing both sides,we get $x+iy = (a+ib)^3$.Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $x+iy = a^

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$-2(a^2+b^2)$

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(A) Given $(x+iy)^{1/3} = a+ib$.Cubing both sides,we get $x+iy = (a+ib)^3$.Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $x+iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$.$x+iy = a^3 + 3a^2bi - 3ab^2 - ib^3$.Grouping real and imaginary parts: $x+iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$.Equating real and imaginary parts: $x = a^3 - 3ab^2$ and $y = 3a^2b - b^3$.Dividing by $a$ and $b$ respectively: $\frac{x}{a} = a^2 - 3b^2$ and $\frac{y}{b} = 3a^2 - b^2$.Therefore,$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (3a^2 - b^2) = a^2 - 3b^2 - 3a^2 + b^2 = -2a^2 - 2b^2 = -2(a^2+b^2)$.

If $(x+iy)^{1/3} = a+ib$ where $x, y, a, b \in R$ and $i = \sqrt{-1}$,then $\frac{x}{a} - \frac{y}{b} = $

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