If (a+ib)^(1/3) = x+iy, prove that 4(x²+y²) = a/x + b/y

By cubing both sides, we see that:
a + bi = (x + iy)^3 = x^3 + 3x^2yi - 3xy^2 - y^3*i = (x^3 - 3xy^2) + (3x^2y - y^3)i.

Comparing the real and imaginary parts gives:
(i) a = x^3 - 3xy^2 and (ii) b = 3x^2y - y^3.

If we divide (i) by x and (ii) by y and add the results, we get:
a/x + b/y = (x^3 - 3xy^2)/x + (3x^2y - y^3)/y
= x^2 - 3y^2 + 3x^2 - y^2
= 4(x^2 - y^2).

If you meant 4(x^2 - y^2) = a/x + b/y, then there you go.

I hope this helps!