I want to solve the following problem.
Verify Bézout's theorem for the curves $(X^{2}-Y)^{2}-X^{5}=0$ and $X^{4}+X^{3}Y-Y^{2}=0$.
Let $\mathcal{C}=\mathrm{V}(F)$ and $\mathcal{C}'=\mathrm{V}(G)$, with $F=(X^{2}-YZ)^{2}Z-X^{5}$ and $G=X^{4}+X^{3}Y-Y^{2}Z^{2}$. We compute the resultant: $$R_{F,G}^{X}=Y^{11}Z^{9}-8Y^{10}Z^{10}-16Y^{9}Z^{11}=$$$$=Y^{9}Z^{9}(Y^{2}-8YZ-16Z^{2})=Y^{9}Z^{9}(Y-(4+4\sqrt{2})Z)(Y-(4-4\sqrt{2})Z)$$
There are four intersection points and their multiplicities must add $20$.
Bézout's theorem says that $$\sum _{P\in P^2} \text{mult}_P(\mathcal C \cap \mathcal C ')=\text{degree}(F)\cdot \text{degree}(G)$$
So it verifies Bèzout's theorem:$$20=\sum _{P\in P^2} \text{mult}_P(\mathcal C \cap \mathcal C ')=\text{degree}(F)\cdot \text{degree}(G)=5\cdot 4 = 20$$
Edit: I computed the resultant in a wrong way, so the result didn't make sense. Now it is ok.