Once you have the cubic equation in $x$, remember that the given point $(-1,3)$ tells you one solution for $x$. How can knowing one solution help you reason about the others?

Write your cubic in the form $x^3 + ax^2 + bx + c = 0$. What is the coefficient of $x^2$, and what does that tell you about $x_1 + x_2 + x_3$? Then use the known root $x_1 = -1$ to find $x_2 + x_3$.

In Desmos, enter y = x^3 - 3x + 1 on one line and y = 2 - x on another. You should see a cubic curve and a straight line.

Click on the points where the line and cubic intersect. Desmos will show three intersection points and their coordinates. Identify the one with $x = -1$ and note the $x$-coordinates of the other two intersection points.

Either write down the two $x$-values and add them by hand, or in a new Desmos expression type their sum (for example, if they are $a$ and $b$, type a + b). The resulting value is the sum of the $x$-coordinates of the other two intersection points.

At intersection points, the $y$-values from both equations are the same, so set the right-hand sides equal:

Move all terms to one side:

So the x-coordinates of intersection points are the solutions (roots) of $x^3 - 2x - 1 = 0$.

The problem tells you that one intersection point is $(-1,3)$, so $x=-1$ is a solution of the equation $x^3 - 2x - 1 = 0$.

Let the three roots (x-coordinates of intersections) be $x_1, x_2, x_3$, with $x_1 = -1$.

For a cubic of the form

the sum of the roots is $x_1 + x_2 + x_3 = -a$ (the negative of the $x^2$ coefficient).

Rewrite the equation in the standard form to read off $a$:

This is the same as $x^3 + 0x^2 - 2x - 1 = 0$, so $a = 0$.

Therefore, the sum of all three roots is

We want $x_2 + x_3$, the sum of the other two intersection x-coordinates. Use

Since $x_1 = -1$, we get

So, the sum of the $x$-coordinates of the other two intersection points is 1.