Weierstrass function: Polynominal with third-periods

Question

Consider a lattice $L=w_1 \mathbb{Z}+w_2\mathbb{Z}$, $w_3=w_1+w_2$ and $w_4=w_1-w_2$.
Let $e_i=\wp(\frac{w_i}{3})$, $1 \leq i \leq 4$ be the "third periods" of $\wp$. Then I have to proof that
$$ 48X^4−24g_2X^2−48g_3X−g_2^2= 48(X−e_1)(X−e_2)(X−e_3)(X−e_4) $$ with $X \in \mathbb{C}$. Here $g_2=g_2(\Omega)=60G_4({\Omega})$ and $g_3=140 G_6 (\Omega)$ are defined via the Eisenstein series.
My attempt: I have tried a similar strategy as in the more known case with $4X^3-g_2X-g_2=4(X-e_1)(X-e_2)(X-e_3)$ with $\textbf{half periods } e_i$. This involved establishing 2 ODE for $(\wp')^2$ and then seeing that they are equal and then plugging in $X$ instead of the Weierstrass function. However, in order to proof the identity $$ (\wp'(z))^2=4(\wp(z)-e_1)(\wp(z)-e_2)(\wp(z)-e_3) $$ you check that both sides of the equations have the same poles and then use liouvilles theorem to see that they are equal up to a constant. However, for the third-periods, I could not establish such a result. This due to the fact that the third-periods do no longer have the property that $\wp(z)-e_i$ has a double zero - as far as I know. Any idea on how to deal with the third-periods? Any help is appreciated!