We have, (x+iy)3 = u + iv
(x+iy)3 = x3+(iy)3+3x2(iy)+3x(iy)2 = x3–y3i+3x2yi–3xy2
= (x3–3xy2)+i(3x2y–y3)
⇒ (x3–3xy2)+i(3x2y–y3) = u + iv
Comparing the real and imaginary parts, we get
u = x3–3xy2, v = 3x2y–y3
u = x(x2–3y2), v = y(3x2–y2)
⇒

u

x

= (x2−3y2) ,

v

y

= 3x2−y2
∴

u

x

+

v

y

= 4x2−4y2 = 4(x2−y2)