Once you set $x^3 - 6x$ equal to $x$, move all terms to one side so that the equation is in the form (polynomial) $= 0$. Then look for a common factor.

After factoring and solving for $x$, remember that on the line $y = x$, the coordinate pairs are $(x,x)$. Match the nonzero solutions to the form $(a,a)$ and think about what $a^2$ must be.

In Desmos, type y = x^3 - 6x on one line and y = x on another. You will see a cubic curve and a straight line.

Click on the points where the graphs intersect. Desmos will show three intersection points; one is at the origin, and the other two are symmetric across the origin.

Look at the coordinates of the nonzero intersection point in the first quadrant, which has the form $(a,a)$. Note the value of $a$ from the graph, then square that value to get $a^2$.

At intersection points, the $y$-values of the two graphs are the same. So set the right sides of the equations equal:

Subtract $x$ from both sides to get a single polynomial equal to zero:

Combine like terms:

Factor out the common factor $x$:

Set each factor equal to zero:

• $x = 0$
• $x^2 - 7 = 0$ (so $x^2 = 7$)

Thus, the $x$-coordinates of the intersection points are $0$ and the two solutions to $x^2 = 7$, which are $x = \pm \sqrt{7}$. Since $y = x$ on the line, these give intersection points $(0,0)$, $(\sqrt{7}, \sqrt{7})$, and $(-\sqrt{7}, -\sqrt{7})$.

The problem tells you the nonzero intersection points are $(a,a)$ and $(-a,-a)$. From the solutions above, these must be $(\sqrt{7}, \sqrt{7})$ and $(-\sqrt{7}, -\sqrt{7})$, so $a = \sqrt{7}$.

Therefore,

The correct answer is $7$.