If $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and vectors $(1,a,a^2)$, $(1,b,b^2)$ and $(1,c,c^2)$ are non-coplanar, then the product $abc$ equals:

The given determinant can be split into two:
$$\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$$
Interchanging columns in the first determinant to match the second:
$$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$$
$$(1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$$
Since the vectors $(1,a,a^2)$, $(1,b,b^2)$, and $(1,c,c^2)$ are non-coplanar, the determinant $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0$.Therefore, $1+abc=0⟹abc=-1$.