If $X + iY$ is a cube root of $x + iy$, then show that $4(X^2 - Y^2) = \frac{x}{X} + \frac{y}{Y}$

Question

I tried expanding $(X + iY)^3$ and equate it with $x + iy$ but I think that is of no use.

Note: Here $i = \sqrt-1$

Answer 1

$$x+iy=(X+iY)^3=X(X^2-3Y^2)+iY(3X^2-Y^2)$$

Equate the real & the imaginary parts assuming $XY\ne0$

Answer 2

Well, we know that $(X+iY)^3 = X^3-XY^2-2XY^2 + i(X^2Y-Y^3+2X^2Y) = x+yi$.

Then, comparing real and imaginary parts, we have:

$$x = X^3-XY^2-2XY^ \implies \frac{x}{X} = X^2-Y^2-2Y^2$$ $$y = X^2Y-Y^3+2X^2Y \implies \frac{y}{Y} = X^2-Y^2+2X^2$$

Summing, we obtain: $\frac{x}{X} + \frac{y}{Y} = 2X^2-2Y^2-2Y^2+2X^2 = 4(X^2-Y^2)$.