I refer to Exercise 2.26 in the Eisenbud-Harris book "3264 and all that". Work over an algebraically closed field of characteristic zero. Let $\sigma:\mathbb{P}^2\times\mathbb{P}^2\rightarrow\mathbb{P}^8$ be the Segre embedding. What is the class of the graph $\Gamma_\sigma\subset\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^8$ in the Chow group $A^8(\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^8)$?
My problem is that I have two approaches to this exercise, and they give different answers! I would appreciate any explanation of which approach is wrong (maybe even both?).
We know $A(\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^8)=\mathbb{Z}[\alpha,\beta,\gamma]/(\alpha^3,\beta^3,\gamma^9)$ where $\alpha,\beta,\gamma$ are the pullbacks of the hyperplane sections of $\mathbb{P}^2,\mathbb{P}^2,\mathbb{P}^8$ respectively.
Answer 1: $\Gamma_\sigma$ is the preimage under $\sigma\times\mathrm{id}:(\mathbb{P}^2\times\mathbb{P}^2)\times\mathbb{P}^8\rightarrow\mathbb{P}^8\times\mathbb{P}^8$ of the diagonal $\Delta_{\mathbb{P}^8}$ of $\mathbb{P}^8$. We know
$$ [\Delta_{\mathbb{P}^8}] = \sum_{i=0}^8\delta^i\gamma^{8-i} $$
where $\delta$ is (the pullback via the other projection of) the hyperplane section of $\mathbb{P}^8$. Since the preimage under $\sigma$ of a hypersurface in $\mathbb{P}^8$ is defined by a bilinear form, we have $\sigma^*\delta=\alpha+\beta$. Hence,
$$[\Gamma_\sigma] = (\sigma\times\mathrm{id})^*[\Delta_{\mathbb{P}^8}] = \sum_{i=0}^4(\alpha+\beta)^i\gamma^{8-i} $$
Answer 2: $\Gamma_\sigma$ is given, in homogeneous coordinates $[x_0,x_1,x_2],[y_0,y_1,y_2],[\{z_{ij}\}_{0\le i,j\le2}]$, as the intersection of 8 hypersurfaces of the form $\{x_iy_jz_{kl} = x_ky_lz_{ij}\}$. Each such hypersurface is given by a trilinear form, so has class $\alpha+\beta+\gamma$. Then
$$[\Gamma_\sigma] = (\alpha+\beta+\gamma)^8 = \sum_{i=0}^4{8\choose i}(\alpha+\beta)^i\gamma^{8-i} $$