Finding Im$f$ when $Re(f) = x^3 - 3xy^2$

Let $f(z) = u(x, y) + iv(x,y)$ and also $f$ is a holomporphic function, where $u(x, y) = x^3-3xy^2$ I want to find $v(x, y)$.

My work so far

$$\frac{\partial u}{\partial x } = 3x^2 - 3y^2 = \frac{\partial v}{\partial y} $$

$$v = \int (3x^2 - 3y^2) dy = 3x^2y - y^3 + \alpha(x)$$

On the other hand

$$\frac{\partial u}{\partial y } = -6xy = -\frac{\partial v}{\partial x} $$

$$v = \int6xy \;dx = 3x^2y + \beta(y)$$

Comparing those two $v$:

$$3x^2y - y^3 + \alpha(x) = 3x^2y + \beta(y)$$

$$\alpha(x) = \beta(y) + y^3$$

And now I got stucked. What exactly can I say about forms of functions $\alpha$ and $\beta$?

Could you please give me a hand in deriving their true form?

The way I liked to think these partial derivative problems is you take $v_x \cup v_y$

Where $v_x$ and $v_y$ are without the constants.

In your case, $v_x \cup v_y$ =$(3x^2y-y^3) \cup (3x^2y)$ =$3x^2y-y^3$

Note: My notations are mathematically flawed. It's just what I used to think it like. It's basically what goes in your head and not how you write down.

Also as you did it, if there was $\alpha(x)$ it would appear when you found $v_x$. Since it didn't hence, $\alpha(x)=0$ and the same way $\beta(y)=-y^3$

Also if you're wondering what is $f(x)$, well most likely it's $f(x)=(x+iy)^3$