For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$

Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?

Yes, this is enough. Basically, if you are trying to solve:
$$\begin{pmatrix}d_1&d_2&d_3\end{pmatrix}A = \begin{pmatrix}y_1&y_2&y_3\end{pmatrix}$$ you are looking for a polynomial $f(x)=d_1+d_2x+d_3x^2$ such that $f(a)=y_1$, $f(d_2)=y_2$, and $f(d_3)=y_3$. 
This we can find by simple interpolation of polynomials - given any $n$ distinct $a_i$ and $n$ distinct $b_i$ we can find a polynomial $p(x)$ of degree at most $n$ such that $p(a_i)=b_i$ for all $i$.
