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?
 A: We use a formulation of collinearity which equates gradients (assuming our points are distinct) $$\frac {y_2-y_1}{x_2-x_1}=\frac {y_3-y_1}{x_3-x_1}$$
This becomes $$\frac {b^3-a^3}{b-a}=\frac {c^3-a^3}{c-a}$$ which leaves us with $$b^2+ab+a^2=c^2+ac+a^2$$ so that $$b^2-c^2=a(c-b)$$
$c\neq b$ so we have $a=-(b+c)$ or $$a+b+c=0$$
Each of the steps can be reversed to prove the converse.
A: We need $$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}=0$$
Now,
$$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}$$
$$=\det\begin{pmatrix} a^3-b^3 & a-b & 0 \\ b^3 & b & 1 \\ c^3-b^3 & c-b &0\end{pmatrix}$$
$$=-(a-b)(b-c)\det\begin{pmatrix} a^2+ab+b^2 & 1 & 0 \\ b^3 & b & 1 \\ c^2+cb+b^2 &1&0\end{pmatrix}$$
$$=-(a-b)(b-c)\{a^2+ab+b^2-(c^2+cb+b^2)\}$$
$$=-(a-b)(b-c)(a^2+ab-bc-c^2)$$
$$=-(a-b)(b-c)\{(a+c)(a-c)+b(a-c)\}$$
$$=(a-b)(b-c)(c-a)(a+b+c)$$
A: Given three distinct points in ${\mathbb{R}^2}$ $P1=(a,a^3)$, $P2=(b,b^3)$ and $P3=(c,c^3)$ for $a,b,c \in \mathbb{R}$, we define ${\vec u},{\vec v} \in \mathbb{R}^2$ such that
$$\vec u = P2-P1=\left( {\begin{array}{*{20}{c}}
  {{b^3} - {a^3}} \\ 
  {b - a} 
\end{array}} \right), \vec v = P3-P1=\left( {\begin{array}{*{20}{c}}
  {{c^3} - {a^3}} \\ 
  {c - a} 
\end{array}} \right)$$
Now, $P1, P2, P3$ are collinear if and only if $||\vec u \times \vec v||=0$, we will show that
$$\begin{gathered}
  ||\vec u \times \vec v|| = ({b^3} - {a^3})(c - a) - ({c^3} - {a^3})(b - a)  \\
  {\text{          }} = (b - a)({b^2} + ba + {a^2})(c - a) - (c - a)({c^2} + ca + {a^2})(b - a)  \\
  {\text{          }} = (b - a)(c - a)({b^2} + ba + {a^2} - {c^2} - ca - {a^2})  \\
  {\text{          }} = (b - a)(c - a)({b^2} + ba - ca - {c^2}) \\ 
   = (b - a)(c - a)[(b - c)(b + c) + a(b - c)]  \\
   = (b - a)(c - a)[(b - c)(b + c + a)]  \\
   = (b - a)(c - a)(b - c)(a + b + c)  \\ 
\end{gathered} $$
since $a+b+c=0$ and $b\ne a, c \ne a, b \ne c$ because $P1,P2,P3$ are distinct, we have $||\vec u \times \vec v||= (b - a)(c - a)(b - c)(a + b + c) = 0$ if and only if $a+b+c=0$.
