Condition for collinearity of points $(a, a^3), (b, b^3), and (c, c^3)$ The following is a statement I have been trying to prove (while solving problem 1.4.26 in Algorithms (4th edition) by Robert Sedgewick).

Show that three points $(a, a^3), (b, b^3), and 
(c, c^3)$ are   collinear if and only if $a + b + c = 0$.

I am aware that the rule for collinearity of three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is -

$x_1(y_2- y_3 ) + x_2(y_3 - y_1 ) + x_3(y_1 - y_2 ) = 0$
  OR
  $(x_2 - x_1) : (x_3 - x_2)  =  (y_2 - y_1) : (y_3 - y_2)$

But I cannot factor out the term $(a+b+c)$ when I apply either of these rules to the points $(a, a^3), (b, b^3), and (c, c^3)$.
Can someone help me with this proof?

We need $$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}=0$$
Now,
$$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}$$
$$=\det\begin{pmatrix} a^3-b^3 & a-b & 0 \\ b^3 & b & 1 \\ c^3-b^3 & c-b &0\end{pmatrix}$$
$$=-(a-b)(b-c)\det\begin{pmatrix} a^2+ab+b^2 & 1 & 0 \\ b^3 & b & 1 \\ c^2+cb+b^2 &1&0\end{pmatrix}$$
$$=-(a-b)(b-c)\{a^2+ab+b^2-(c^2+cb+b^2)\}$$
$$=-(a-b)(b-c)(a^2+ab-bc-c^2)$$
$$=-(a-b)(b-c)\{(a+c)(a-c)+b(a-c)\}$$
$$=(a-b)(b-c)(c-a)(a+b+c)$$
