What is the difference between the two real numbers that satisfy this equation? What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form


I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.
We have
$$(x^2-1)(x-2) = (x^2-9)(x+2) \implies x^3-2x^2-x+2 = x^3+2x^2-9x-18$$
which simplifies to
$$4x^2-8x-20 = 0 \implies x^2-2x-5 = 0 \implies (x-1)^2 = 6$$
Hence, the roots are $1\pm\sqrt6$, which means that the difference between the roots is $2\sqrt6$.
$$(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$$
$$\frac{(x+2)}{(x-2)}=\frac{x^2-1}{x^2-9}$$
$$1+\frac{4}{x-2}=1+\frac{8}{x^2-9}$$
$$4x^2-36=8x-16$$
$$x^2-2x-5=0$$
Now difference of roots is $$\frac{\sqrt{D}}{a}=\sqrt{24}$$