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Factorise $9x-x^3$ completely.

It's simple but I'm never seem to get it right; I've got $(x-1)(-x+9)x$.

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$$9x-x^3 = x(9-x^2)=x(3-x)(3+x)$$

where in the last step I have used the very important and useful "difference of two squares" identity:

$$ (A-B)(A+B)=A^2-B^2 $$

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Notice that $9 - x^2$ is the product of $(3 - x)(3+x)$

$$9 x - x^3 =x (9 - x^2) = x(3-x)(3+x)$$

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