If (x + iy)1/3 = a + ib, then

(d) none of these

$(x+i y)^{\frac{1}{3}}=a+i b$

Cubing on both the sides, we get:

$x+i y=(a+i b)^{3}$

$\Rightarrow x+i y=a^{3}+(i b)^{3}+3 a^{2} b i+3 a(i b)^{2}$

$\Rightarrow x+i y=a^{3}+i^{3} b^{3}+3 a^{2} i b+3 i^{2} a b^{2}$

$\Rightarrow x+i y=a^{3}-i b^{3}+3 a^{2} i b-3 a b^{2} \quad\left(\because i^{2}=-1, i^{3}=-i\right)$

$\Rightarrow x+i y=a^{3}-3 a b^{2}+i\left(-b^{3}+3 a^{2} b\right)$

$\therefore x=a^{3}-3 a b^{2}$ and $y=3 a^{2} b-b^{3}$

or,$\frac{x}{a}=a^{2}-3 b^{2}$ and $\frac{y}{b}=3 a^{2}-b^{2}$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=a^{2}-3 b^{2}+3 a^{2}-b^{2}$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=4 a^{2}-4 b^{2}$