$$
\begin{aligned}
T(n) &=T(n-2)+(n-1)^{2}+n^{2} \\
&=T(n-3)+(n-2)^{2}+(n-1)^{2}+n^{2} \\
&=T(n-4)+(n-3)^{2}+(n-2)^{2}+(n-1)^{2}+n^{2} \\
T(n) &=T(0)+1^{2}+2^{2}+3^{2}+\cdots+n^{2} \\
1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
\end{aligned}
$$
so
$$
T(n)=T(0)+\frac{n(n+1)(2 n+1)}{6}
$$
so we conclude that $T(n)=O\left(n^{3}\right)$, in asymptotic notation.
