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?
 A: Note that $(a^2+ab)-(c^2+cb)=(a-c)(a+c)+b(a-c)=(a-c)(a+b+c)$
A: 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
$$\color{red}{(b-a)(c-a)(c-b)}\color{blue}{(a+b+c)}.
$$
A: $$\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}$$
$$=\begin{vmatrix}1&1-1\\c^2+cb+b^2&a^2+ab+b^2-(c^2+cb+b^2)\end{vmatrix}$$
$$=\begin{vmatrix}1&0\\c^2+cb+b^2&a^2-c^2+b(a-c)\end{vmatrix}$$
$$=a^2-c^2+b(a-c)$$
$$=(a-c)(a+c+b)$$
A: You can get it immediately:
$$\sum_{cyc}(a^3c-a^3b)=(a+b+c)(a-b)(b-c)(c-a)$$
because for $a=b$ or $a=c$ or $b=c$ our determinant is equal to zero, which gives
$$K(a+b+c)(a-b)(b-c)(c-a)$$ and it's enough to check the coefficient before $a^3c$, which is $1$, which gives $K=1$.
A: Here is a geometrical interpretation of the formula :
$$(a-b)(b-c)(c-a)(a+b+c).\tag{1}$$
I use the term "interpretation" because I have not written here a complete proof but rather an inductive way to obtain (1).
Indeed, if the presence of factors $(a-b), (b-c), (c-a)$ look very natural (the determinant is zero if at has 2 identical columns), this is apparently not the case for factor $(a+b+c)$.
Here is a context giving a natural interpretation to this factor.
Reminder (see Theorem in https://proofwiki.org/wiki/Area_of_Triangle_in_Determinant_Form ) : Let $A,B,C$ be $3$ points in the plane.
$$\begin{vmatrix}1&1&1\\x_A&x_B&x_C\\y_A&y_B&y_C\end{vmatrix}=2 \times area(ABC)$$
(being understood that, is $A,B,C$ are all different, this determinant is zero iff $A,B,C$ are aligned).
Here, we take points $A,B,C$ on the curve $\Gamma$ with equation $y=x^3$. If they are distinct, they are aligned iff their abscissas are such that :
$$x_A+x_B+x_C=0$$
which is rather convincing when one looks at curve $\Gamma$ (Fig. 1) :

Fig. 1 : the sum of abscissas of aligned points $A,B,C$ is $-1.5+0.5+1=0$.
Now, the proof : the abscissas of intersection points of :
$$\begin{cases}y&=&x^3\\y&=&ax+b\end{cases} $$
verify 
$$x^3-\underbrace{0}_S x^2-ax-b=0.$$
The sum of roots $S$ (coefficient of $-x^2$ according to Viète's formulas) is thus zero.
A: 1) By inspection : The determinant $= 0$ for $a=b$; $a=c$, and $b=c$ (Why?).
Hence $\pm (c-b)$, $\pm(a-b)$, and $\pm (a-c)$ are factors.
You got $(c-b)(a-b)[a^2+ab -(c^2+cb)]$.
Hence $\pm (a-c)$ is a factor of $[a^2+ab -(c^2+cb)]$.
$a^2-c^2+b(a-c)=$
$ (a-c)(a+c)+b(a-c)=$
$(a-c)(a+b+c).$
