Cubic Intersections!

Given $\begin{cases} y = 2x^3-4x+2 & ...(1) \\ y = x^3+2x-1 & ...(2) \end{cases}$ . The three points of intersection satisfy $(1)-(2): x^3 - 6x + 3=0$ . At the point of intersect $y$ is given by:

$\begin{aligned} (1): \quad y & = 2x^3-4x+2 \\ & = 2({\color{#3D99F6}x^3-6x+3}) + 8x - 4 & \small \color{#3D99F6} \text{At points of intersection: }x^3-6x+3=0 \\ \implies y & = 8x-4 \end{aligned}$

Similarly,

$\begin{aligned} (2): \quad y & = x^3+2x-1 \\ & = {\color{#3D99F6}x^3-6x+3} + 8x - 4 & \small \color{#3D99F6} \text{At points of intersection: }x^3-6x+3=0 \\ \implies y & = 8x-4 \end{aligned}$

Therefore, the three points of intersection are on a straight line $y = 8x-4$ and the gradient of the line is $\boxed 8$ .

Finding the approximate points of intersection, it's easy to guess the line passing through all three points of intersection is $y=8x-4$ , but why?

If you set the two equations equal to each other you get $2x^{3}-4x+2=x^3+2x-1$ which simplifies to $x^{3}-6x+3=0$ which has the same solutions for $x$ , but the problem is their closed form is the crazy cubic type. Yuck.

Ok, how about we try to make use of the probable line $y=8x-4$ but subtracting it from the original two equations?

The first becomes $y=2x^{3}-12x+6$ which is just double the above, so it has those same crazy zeros.

The second becomes $y=x^{3}-6x+3$ which is the same thing. Same crazy zeros.

So everything works out nicely with that slope of $\boxed{8}$