The recurrences of the form
$$T(n)=T(n-1)+f(n),\\T(0)=T_0$$ are solved by
$$T(n)=T_0+\sum_{k=1}^n f(n)$$
as you can check by induction.
(Because $T(n)=T(n-1)+f(n)=T(n-2)+f(n-1)+f(n)=T(n-3)+f(n-2)+f(n-1)+f(n)=\cdots$.)
From the above formula, we can obtain the asymptotic expression in an empirical way.
As the terms $n^2$ go growing, one can assume that the average resembles $n^2$ and we try the ansatz $\dfrac{T(n)}n=an^2$.
This leads to $$an^3=a(n-1)^3+n^2=an^3+(1-3a)n^2+3an-a,$$
which simplifies most when $a=\dfrac13$: $$0\sim n-\frac13.$$
As this residual is of a lower degree, we conclude
$$T(n)\sim\frac{n^3}3=\Theta(n^3).$$
Another method is to replace the sum by an integral.