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

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(B) Given $(x + iy)^{1/3} = a + ib$. Cubing both sides,we get: $x + iy = (a + ib)^3$ $x + iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$ $x + iy = a^3 + 3a^

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$4(a^2 - b^2)$

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(B) Given $(x + iy)^{1/3} = a + ib$. Cubing both sides,we get: $x + iy = (a + ib)^3$ $x + iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$ $x + iy = a^3 + 3a^2bi - 3ab^2 - ib^3$ $x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$ 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 $\frac{y}{b} = 3a^2 - b^2$ Adding these two: $\frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2) = 4a^2 - 4b^2 = 4(a^2 - b^2)$

If $(x + iy)^{1/3} = a + ib$,then $\frac{x}{a} + \frac{y}{b}$ is equal to

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