What is $q(x)$ and $r(x)$ when $(x^2-6x+9)q(x) + r(x) = x^3 -27$?

Question

I just failed this question on a test, so I would please like to get some feedback on where my thinking was wrong.

I need help with determining $q(x)$ and $r(x)$ when: $$(x^2-6x+9)q(x) + r(x) = x^3 -27$$

I know that: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

In this case $a=x$ and $b=\pm3$. By using $+3$ you get the calculation: $$(x^2-6x+9)(x-3) = x^3-9x^2+27x -27$$

To get the final expression: $x^3 -27$ you have to set $r(x)$ to be: $9x^2-27x$.

But this was wrong. Where is my line of thinking incorrect?

Thank you kindly for your help!

Answer 1

Long division!

$$\require{enclose} \begin{array}{rll} x\;+6\qquad\qquad\quad\\[-3pt] x^2-6x+9\enclose{longdiv}{x^3\qquad\qquad\;\;-27}\\ \underline{x^3-6x^2+9x\phantom{\qquad\,}}\\[-3pt] 6x^2-\phantom{0}9x-27\\[-3pt] \underline{6x^2-36x+54\;}\\[-3pt] 27x-81 \end{array}$$

So $q(x) = x + 6$ and $r(x) = 27 x - 81$.