After you expand $f(x+2)$ and $f(x-2)$, carefully subtract the second from the first. Pay attention to how the $x^3$ and $x$ terms behave.

Once you have $g(x)$ as a simpler polynomial, set $g(x)=0$ to find the $x$-intercepts. If that equation is quadratic, recall that for $Ax^2+Bx+C=0$, the product of the two solutions is $C/A$—you can use this directly to find $ab$.

In a Desmos expression line, type f(x) = x^3 - 9x to define the original function.

In a new line, type g(x) = f(x+2) - f(x-2); Desmos will plot the graph of $g$.

Click or tap on the points where the graph of $g$ crosses the $x$-axis. Desmos will display the coordinates of these intercepts; note the two $x$-values (these are $a$ and $b$).

In a new expression line, type the product of the two $x$-values you found (for example, something like (first_x_value)*(second_x_value)). The resulting number shown by Desmos is the value of $ab$.

We are given $f(x)=x^3-9x$ and $g(x)=f(x+2)-f(x-2)$.

First find $f(x+2)$:

Now find $f(x-2)$:

Now compute $g(x)$ by subtracting:

Distribute the minus sign in the second parentheses and combine like terms:

So $g(x)=12x^2-20$, a quadratic function.

The $x$-intercepts occur where $g(x)=0$.

So the $x$-coordinates $a$ and $b$ satisfy

In general, for a quadratic equation written as

with solutions (roots) $r_1$ and $r_2$, the product of the roots is

Here, the roots of $12x^2 - 20 = 0$ are exactly $a$ and $b$, so $ab$ equals $C/A$ for this quadratic.

For $12x^2 - 20 = 0$, we have $A=12$ and $C=-20$, so

Thus, the value of $ab$ is $-\frac{5}{3}$.