Show that for prime $p$, $\sum_{c=1}^{p-1}(\frac{c^2-1}{p})(\frac{c^2-2}{p})\sum_{b=1}^{p-1}(\frac{b^2-c^2}{p})(\frac{b^2-2}{p})\ll p$

Question

Here $\left(\frac{\cdot}{\cdot}\right)$ is the Legendre symbol. I hope that \begin{align}\left|\sum_{c=1}^{p-1}(\frac{c^2-1}{p})(\frac{c^2-2}{p})\sum_{b=1}^{p-1}(\frac{b^2-c^2}{p})(\frac{b^2-2}{p})\right|\\ \leq \left|\sum_{c=1}^{p-1}(\frac{c^2-1}{p})(\frac{c^2-2}{p})\sum_{b=1}^{p-1}(\frac{b^2-1}{p})(\frac{b^2-2}{p})\right|.\end{align} Now both $\sum_{c=1}^{p-1}(\frac{c^2-1}{p})(\frac{c^2-2}{p})$ and $\sum_{b=1}^{p-1}(\frac{b^2-1}{p})(\frac{b^2-2}{p})$ are $O(\sqrt{p})$, by Weil estimate for character sum. So the required sum is $O(p)$. But here I am doubtful about the first inequality.