Hi I need to find the values of x for which

$\displaystyle \frac{x^{3}+5x-12}{x-3} > 4 $

Now I could - 4(x-3)/(x-3) from the RHS to the LHS, which is what the markscheme does, but why cant I do this:

$\displaystyle x^{3}+5x-12> 4(x-3) $

$\displaystyle x^{3}+5x-12> 4x-12 $

$\displaystyle x^{3}+5x> 4x $

$\displaystyle x^{3}+x $

that only gives me x = 0 and i as opposed to the other way which gives me 0,3,i

Why doesn't the stupid math work?

Thanks