<s>[INST] Prove equation $(ad-bc)(a-c)^2 = (b-d)^3$, if polynomials has common root 
$$\begin{split}
 W(x) &= x^3 + ax + b \wedge a,b \in \mathbb{R} 
  &\wedge \mathbb{D}_W &= \mathbb{R}\\
 G(x) &= x^3 + cx + d \wedge c,d \in \mathbb{R}
  &\wedge \mathbb{D}_G &= \mathbb{R}
 \end{split}
 $$ Prove:
  $$ \left(\exists p \in \mathbb{R}\right)\left(W(p) = 0 = G(p)\right)
\Longrightarrow \left((ad-bc)(a-c)^2 = (b-d)^3\right)$$

I cannot prove it. It's obvious if $p = 0$ cause $d=b=0$. But otherwise I don't see solution. How can I prove it?

I noticed that $pa+b =cp+d$, so $(b-d) = p(c-a)$. Therefore 
$(b-d)^3 = 
%p^2(c-a)^2(b-d)=
(a-c)^2 \cdot p^2(b-d)$.
Now I just have to show, that $p^2(b-d) = (ad-bc)$.
I tried ($b \neq d)$ show that:
$$ \frac{ad}{b-d} - \frac{bc}{b-d} = p^2$$
From first observation, if $c \neq a$, $p^2 = \left( \frac{b-d}{c-a} \right)^2$, but it's now worthless.I don't see next step. Can you show me, how I should finish that proof?
 [/INST] Let $y$ be the common root. We then have
$$y^3 + ay+b = 0 \text{ and }y^3+cy+d = 0$$
This means we have
$$ay+b = cy+d \implies y = \frac{d-b}{a-c}$$
Hence,
$$\left(\frac{d-b}{a-c}\right)^3 + a \left(\frac{d-b}{a-c}\right) + b = 0$$
This gives us
$$(d-b)^3 + a(d-b)(a-c)^2 + b(a-c)^3 = 0 \implies (b-d)^3 = (a-c)^2(ad-ab+ba-bc)$$
which simplifies to what you are after.
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