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

$(a)\;4(x^{2}-y^{2})\qquad(b)\;4(y^{2}-x^{2})\qquad(c)\;4(y-x)\qquad(d)\;4(x-y)$

Answer : $\;4(x^{2}-y^{2})$
Explanation :
$\;(x+iy)^{3}=u+iv\;$
$x^{3}+(iy)^{3}+3\;.x\;.(iy) (x+iy) =u+iv$
$x^{3} +i^{3}y^{3}+3x^{2}iy + 3xi^{2}y^{2} = u+iv$
$x^{3} -iy^{3} +3x^{2}yi -3xy^{2} = u+iv$
$(x^{3} -3xy^{2}) + i (3x^{2}y - y^{3}) = u+iv$
On equating real and imaginary parts , we obtain
$u = x^{3} -3xy^{2} \;v = 3x^{2}y-y^{3}$
$\; \large\frac{u}{x} + \large\frac{v}{y} = \large\frac{x^{3} -3xy^{2} }{x} + \large\frac{3x^{2}y-y^{3}}{y}$
$= \large\frac{x(x^{2}-3y^{2})}{x} + \large\frac{y(3x^{2}-y^{2})}{y}$
$= x^{2} - 3y^{2} + 3x^{2} -y^{2} = 4x^{2} - 4y^{2}$
$= 4(x^{2} -y^{2})$
Therefore , $\; \large\frac{u}{x} + \large\frac{v}{y} = 4(x^{2}-y^{2})$