Factorise $9x-x^3$ completely.
It's simple but I'm never seem to get it right; I've got $(x-1)(-x+9)x$.
$$9x-x^3 = x(9-x^2)=x(3-x)(3+x)$$
where in the last step I have used the very important and useful "difference of two squares" identity:
$$ (A-B)(A+B)=A^2-B^2 $$
Notice that $9 - x^2$ is the product of $(3 - x)(3+x)$
$$9 x - x^3 =x (9 - x^2) = x(3-x)(3+x)$$