<s>[INST] How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
 [/INST] $$x^3-4x^2+5x-6 = (x-3)(x^2-x+2)$$
$$x^2-x-6 = (x-3)(x+2)$$
so the equation becomes: $$\frac{x^2-x+2}{x+2} =4 $$
$$x^2-x+2=4x+8$$
$$x^2-5x-6=0$$
$$(x-6)(x+1)=0$$
$x=6$ or $-1$
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