<s>[INST] Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$ 
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$

My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
 [/INST] Given a $4\times 4$ Vandermonde matrix
$$
\left[\begin{matrix}
\color{red}1&\color{red}1&\color{red}1&1\\\color{red}a&\color{red}b&\color{red}c&d\\ a^2&b^2&c^2&\color{blue}{d^2}\\\color{red}{a^3}&\color{red}{b^3}&\color{red}{c^3}&d^3
\end{matrix}\right],
$$ note that the desired determinant is the minor of $d^2$. Considering Laplace expansion, it appears as a minus(-) coefficient of $d^2$ in the Vandermonde determinant
$$\begin{align*}
&\color{red}{(b-a)(c-a)(c-b)}(d-a)(d-b)(d-c)\\=&\color{red}{(b-a)(c-a)(c-b)}(d^3-\color{blue}{(a+b+c)}d^2+\cdots).
\end{align*}$$ Thus we obtain the desired determinant
$$\color{red}{(b-a)(c-a)(c-b)}\color{blue}{(a+b+c)}.
$$
</s>