If z = x + iy,{z^{1/3}} = a - ib and frac{x}{ | vedclass

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(B) Given that $(x + iy)^{1/3} = a - ib$.By cubing both sides,we get $x + iy = (a - ib)^3$.Expanding the right side using $(a - b)^3 = a^3 - 3a^2b + 3

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$4$

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(B) Given that $(x + iy)^{1/3} = a - ib$.By cubing both sides,we get $x + iy = (a - ib)^3$.Expanding the right side using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we have:$x + iy = a^3 - 3a^2(ib) + 3a(ib)^2 - (ib)^3$$x + iy = a^3 - 3ia^2b - 3ab^2 + ib^3$$x + iy = (a^3 - 3ab^2) + i(b^3 - 3a^2b)$.Equating real and imaginary parts:$x = a^3 - 3ab^2$ and $y = b^3 - 3a^2b$.Now,$\frac{x}{a} = a^2 - 3b^2$ and $\frac{y}{b} = b^2 - 3a^2$.Calculating $\frac{x}{a} - \frac{y}{b}$:$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (b^2 - 3a^2)$$\frac{x}{a} - \frac{y}{b} = a^2 - 3b^2 - b^2 + 3a^2 = 4a^2 - 4b^2 = 4(a^2 - b^2)$.Comparing this with $k(a^2 - b^2)$,we get $k = 4$.

If $z = x + iy$,${z^{1/3}} = a - ib$ and $\frac{x}{a} - \frac{y}{b} = k({a^2} - {b^2})$,then the value of $k$ is equal to:

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