If z= x- iy and z^3 = p+IQ then x÷p+ y÷q÷p^2 + q^2 is equal to

Z=x+iy and z¹/³=p+iq
∴, (x+iy)¹/³=p+iq
Cubing both sides,
x+iy=(p+iq)³
or, x+iy=p³+3p²iq+3p(iq)²+(iq)³
or, x+iy=p³+3ip²q-3pq²-iq³  [i²=-1]
or, x+iy=(p³-3pq²)+i(3p²q-q³)
Equating the real and imaginary parts from both sides,
x=p³-3pq² and y=3p²q-q³
∴, (x/p+y/q)/(p²+q²)
=[(p³-3pq²)/p+(3p²q-q³)/q]/(p²+q²)
=(p²-3q²+3p²-q²)/(p²+q²)
=(4p²-4q²)/(p²+q²)
=4(p²-q²)/(p²+q²)