Collinear Points on a Cubic

Let the point R be represented as $R(x_0 , x_0^3 - 3x_0 + 2)$ and the point Q as $Q(\frac{x_0 - 2}{2}, \frac{x_0^3 - 3x_0 + 2}{2}).$ If Q is the midpoint of |PR|, then we must satisfy:

$\frac{x_0^3 - 3x_0 +2}{2} = (\frac{x_0-2}{2})^3 - 3(\frac{x_0-2}{2}) + 2;$

or $\frac{x_0^3 -3x_0 +2}{2} = \frac{(x_0-2)^{3}}{8} - 3(\frac{x_0-2}{2}) + 2;$

or $3x_0^3 + 6x_0^2 - 12x_0 - 24 =0;$

or $x_0^3 + 2x_0^2 - 4x - 8 = 0;$

or $(x_0 - 2)(x_0+2)^2 = 0$

or $x_0 = \pm 2.$

Since P and R are distinct points, we must only admit $x_0 = 2 \Rightarrow Q(0,2)$ and $R(2,4).$ Hence the quantity $|PR|^2$ equals:

$|PR|^2 = (2-(-2))^2 +(4-0)^2 = 2 \cdot 4^2 = \boxed{32}.$

It is not too hard to see $Q=(0,2)$ , the point of symmetry of the cubic. (Try to prove this, ;))

Thus, $R=(2,4)$ , so $PR=\sqrt{32}$ . Therefore, $PR^2=32$ .