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?

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.
