<s>[INST] Long Division of Proper Polynomials I took the derivative of function, yielding the result: $$\frac{-3x^4 - 4x^3 - 3x^2 + 18x + 6}{x^6 - 6x^3 + 9}$$ I know this to be correct and wanted to simplify the answer.
I've never attempted long division on an expression in which the highest order is in the denominator before but attempted:
$$
\require{enclose}
\begin{array}{r11}
   -3x^{-2} - 4x^{-3} - 3x^{-4} ...  \\[-3pt]
   x^6 - 6x^3 + 9 \enclose{longdiv}{-3x^4 - 4x^3 - 3x^2 + 18x + 6}\kern-.2ex \\[-3pt]
      \underline{-3x^4-18x+27x^{-2} } \\[-3pt]
      -4x^3-3x^2+36x+6 \\[-3pt]
      \underline{-4x^3-24-36x^{-3}} \\[-3pt]
      -3x^2+36x-18-+27x^{-2}-36x^{-3} \\[-3pt]
      ...
  \end{array}
$$
It seems that for every term I eliminate, I gain another of a lower order. Is there a problem with my method, do these two expressions not divide cleanly? 
How can one tell whether the division of polynomials is possible?
 [/INST] Note that the denominator is $(x^3-3)^2$, so you can try dividing by $x^3-3$. If you are looking for an exact factor there is a sneaky trick here - you can put $x^3=3$ to test divisibility, and this gives $-9x-12-3x^2+18x+6\neq 0$.
But you can still divide the numerator by $x^3-3$ to obtain $$-3x^4-4x^3-3x^2+18x+6=-(x^3-3)(3x+4)-3x^2+9x-6$$ which leaves your answer in a kind of partial fraction form as $$-\frac{3x+4}{x^3-3}-\frac{3x^2-9x+6}{(x^3-3)^2}=-\frac{3x+4}{x^3-3}-3\frac{(x-1)(x-2)}{(x^3-3)^2}$$
Note also that you can factorise $x^3-3$. If you take $t$ as the real cube root of $3$ then you have the real factorisation $(x-t)(x^2+tx+t^2)$ and if further $\omega^2+\omega+1=0$ so that $\omega$ is a complex cube root of unity then the factorisation can be completed as $(x-t)(x-\omega t)(x-\omega^2 t)$. You can use these to get a breakdown into more conventional partial fractions.
Whether any of this is useful to your purpose will depend on what you are trying to achieve.
</s>