I want to compute cohomology ring of $\mathbb{CP}^2\#\bar{\mathbb{CP}^2}$. I know that as a module they are just $H^*(\mathbb{CP}^2)\oplus H^*(\mathbb{CP}^2)$ due to M-V arguement.
Now, let $x\in H^2(\mathbb{CP}^2)$ and $y\in H^2(\bar{\mathbb{CP}^2})$ be the generators of the cohomology ring respectively. Since $\mathbb{CP}^2$ is oreintable hence $\mathbb{CP}^2\#\bar{\mathbb{CP}^2}$ is also orientable.
I have seen that $H^*(\mathbb{CP}^2\#\bar{\mathbb{CP}^2})=\frac{\mathbb{Z}x\oplus\mathbb{Z}y}{<x^3,y^3,x^2+y^2,xy>}$ as a graded polynomial ring with $|x|,|y|=2$.
Here, I understand why $x^2+y^2=0$, as the fundamental class of $[\mathbb{CP}^2\#\bar{\mathbb{CP}^2}]=[\mathbb{CP}^2]+[\bar{\mathbb{CP}^2}]$, where $[\bar{\mathbb{CP}^2}]=-[\mathbb{CP}^2]$ and hence $$(x^2+y^2)([\mathbb{CP}^2\#\bar{\mathbb{CP}^2}])=(x^2+y^2)([\mathbb{CP}^2]+[\bar{\mathbb{CP}^2}])=1-1=0$$
But I cant figure out how $x\cup y=0$.