Find the $4 \times 4$ Vandermonde determinant I'm currently doing an exercise question from the textbook. The question is:

The goal of this problem is to find the $4 \times 4$ Vandermonde determinant.
$V_4 = \begin{bmatrix}1&a&a^2&a^3\\1&b&b^2&b^3\\1&c&c^2&c^3\\1&x&x^2&x^3\end{bmatrix}$.
(a). Explain why $V_4$ is a cubic polynomial in the variable $x$.
(b). Find three possible values $r_1, r_2, r_3$ for $x$ that make $V_4$ equal to $0$. These are the roots of $V_4$ as a polynomial in $x$.
(c). Explain why $V_4=A(x-r_1)(x-r_2)(x-r_3)$ for some value $A$, and show that the value of $A$ is the $3\times3$ Vandermonde determinant $(b-a)(c-a)(c-b)$.
The $3\times3$ Vandermonde matrix is $V_3 = \begin{bmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{bmatrix}$.
(d). Finally, write down a formula for $V_4$ in terms of $a, b, c, x$.

For question (a), I understand that $V_4$ is a cubic polynomial in the variable $x$ because the $x's$ are on the same row and they won't multiply each other. The highest degree is $3$. Therefore, the determinant is a cubic polynomial.
For question (b), three possible values $r_1, r_2, r_3$ are $a, b, c$. Because if $V_4 =0$, that means the matrix is non-invertible, which then means the rows are not linearly independent. If $x=a$ or $x=b$ or $x=c$, then the rows will cancel each other out and result in a zero row which then leads to the determinant being $0$.
However, I don't really understand $(c)$ and $(d)$. Can someone give me some guidance?
 A: Answering my own question in a different way from
math.stackexchange.com/a/699339/42969
because it's easier to understand (for me and maybe for some other people).
(a). The determinant is a cubic polynomial because if we calculate the determinant by cofactor expansion across the last row, we have
$V_4= -1 \begin{vmatrix}a&a^2&a^3\\b&b^2&b^3\\c&c^2&c^3\end{vmatrix} + x\begin{vmatrix}1&a^2&a^3\\1&b^2&b^3\\1&c^2&c^3\end{vmatrix} - x^2\begin{vmatrix}1&a&a^3\\1&b&b^3\\1&c&c^3\end{vmatrix} + x^3\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$. This shows that the highest degree of this polynomial is $3$, and the coefficients of these variables are cofactors $C_41, C_42, C_43$ and $C_44$ which doesn't contain variable $x$.
(b). Three possible values $r_1, r_2, r_3$ for $x$ to make $V_4$ equal to $0$ are $a, b, c$. Because for $V_4$ to be $0$, the Vandermonde matrix has to be non-invertible, which means the rows have to be not independent. If $x=a$, $x=b$, or $x=c$, then there will be same rows which makes the matrix not independent.
(c). Since $V_4$ is some cubic polynomial, it can be factored as a product of three linear terms $V_4=A(x-r_1)(x-r_2)(x-r_3)$ for some value $A$ which is the coefficient of the cubic term. From my answer in (a), it shows that $A$ is the determinant of the $3\times3$ Vandermonde matrix.
(d). The formula for $V_4$ in terms of $a, b, c, x$ is $V_4=(b-a)(c-a)(c-b)(x-a)(x-b)(x-c)$.
