A cubic factor of a nine degree polynomial The polynomial $x^3−3x^2+4x−1$ is a factor of $x^9+px^6+qx^3+r$. Find $p+q+r$. How do I start with the problem?
 A: I have a feeling there may be a smart way of doing this question, but here is the straightforward way.
If you (carefully!) divide $x^9+px^6+qx^3+r$ by $x^3-3x^2+4x-1$, you will find that you get a remainder of
$$(-5p+3q-63)x^2+(-11p-4q+190)x+(4p+q+r-54)\ .$$
Setting the coefficients of this quadratic to zero and solving the system finds the values of $p,q,r$.
Hover below to reveal the answer.  (BUT please try it yourself first!)

 Answer: $p=6$, $q=31$, $r=-1$, so $p+q+r=36$.

A: Let $\omega=1^{1/3}$ be a complex cube-root of $1$.  Then, if $a(x)$ is the cubic, and $b(x)$ is the degree-nine polynomial, then $b(x)=a(x)a(\omega x)a(\omega^2x)$.  If you expand that, I think you should get the same as the others.
Using $a(x)=(x-1)^3+x$, 
$$p+q+r=b(1)-1=a(1)a(\omega)a(\overline\omega)-1\\=1(-0.5+3.5\sqrt{3}i)(-.5-3.5\sqrt{3}i)-1=36$$
A: As David wrote, the division must be done very carefully. To avoid it, we could use a real brute force approach writing $$\Delta=(x^9+px^6+qx^3+r)-(x^3−3x^2+4x−1)\sum_{i=0}^6 a_ix^i=0$$ This should give $$\Delta=(a_0+r)+(a_1-4 a_0) x+(3 a_0-4 a_1+a_2) x^2+ (-a_0+3 a_1-4
   a_2+a_3+q)x^3+(-a_1+3 a_2-4 a_3+a_4) x^4+(-a_2+3 a_3-4 a_4+a_5) x^5+
   (-a_3+3 a_4-4 a_5+a_6+p)x^6+(-a_4+3 a_5-4 a_6) x^7+(3 a_6-a_5) x^8+(1-a_6)
   x^9=0$$ Now, we cancel the coefficient of each power of $x$ starting from the highest. This will give $a_6=1,a_5=3,a_4=5,a_3=p+4,a_2=3p-5,a_1=5p-26,a_0=-r$. After this quite laborious task, what is left is $$\Delta= (4 p+q+r-54)x^3+(-17 p-3 r+99)x^2 + (5 p+4 r-26)x=0$$ and again we cancel the coefficients to get three linear equations for the three unknowns $p,q,r$ and the solutions as already given in the previous answers.
A: Let $f(x) = x^3-3x^2+4x-1 $ Because the $ 9^\text{th}\ $ degree polynomial only have terms of multiples of $ 3\ $. I consider that it is the root to $ f \left ( \sqrt[3]{x} \right ) = 0 $
We have
$ \begin{eqnarray}
f \left ( \sqrt[3]{x} \right ) & = & 0 \\
x - 3x^{2/3} + 4x^{1/3} - 1 & = & 0 \\
(x-1) & = & x^{1/3} \cdot (3x^{1/3} - 4) \\
(x-1)^3 & = & x( 27x - 64 - 36x^{1/3} (3x^{1/3} - 4)) \\
& = & x(27x- 64 - 36(x-1)) \\
x^3 + 6x^2 + 31x - 1 & = & 0 \\
\end{eqnarray} $
Replace $ x $ with $ x^3 $ and compare, we have 

$ p = 6, q = 31, r = -1 $

