If $(x+iy)^{3}=u+iv$,then show that: $\frac{u}{x}+\frac{v}{y}=4(x^{2}-y^{2})$

(A) Given $(x+iy)^{3}=u+iv$.
Expanding the left side using $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$:
$(x+iy)^{3} = x^{3}+(iy)^{3}+3(x)(iy)(x+iy) = u+iv$
$x^{3}+i^{3}y^{3}+3x^{2}yi+3xy^{2}i^{2} = u+iv$
Since $i^{2}=-1$ and $i^{3}=-i$:
$x^{3}-iy^{3}+3x^{2}yi-3xy^{2} = u+iv$
Grouping real and imaginary parts:
$(x^{3}-3xy^{2}) + i(3x^{2}y-y^{3}) = u+iv$
Equating real and imaginary parts:
$u = x^{3}-3xy^{2}$ and $v = 3x^{2}y-y^{3}$
Now,evaluate $\frac{u}{x}+\frac{v}{y}$:
$\frac{u}{x} = \frac{x^{3}-3xy^{2}}{x} = x^{2}-3y^{2}$
$\frac{v}{y} = \frac{3x^{2}y-y^{3}}{y} = 3x^{2}-y^{2}$
Adding these results:
$\frac{u}{x}+\frac{v}{y} = (x^{2}-3y^{2}) + (3x^{2}-y^{2})$
$= 4x^{2}-4y^{2} = 4(x^{2}-y^{2})$
Hence,proved.