jmc

Let $\alpha$ be a root of $x^3-3x^2+4x-1=0,$ so ${\alpha }^3=3{\alpha }^2-4\alpha +1.$ Then $${\alpha }^4=3{\alpha }^3-4{\alpha }^2+\alpha =3(3{\alpha }^2-4\alpha +1)-4{\alpha }^2+\alpha =5{\alpha }^2-11\alpha +3.$$Hence, $$\begin{array}{rl}{\alpha }^6& =(3{\alpha }^2-4\alpha +1{)}^2\\ & =9{\alpha }^4-24{\alpha }^3+22{\alpha }^2-8\alpha +1\\ & =9(5{\alpha }^2-11\alpha +3)-24(3{\alpha }^2-4\alpha +1)+22{\alpha }^2-8\alpha +1\\ & =-5{\alpha }^2-11\alpha +4,\end{array}$$and $$\begin{array}{rl}{\alpha }^9& ={\alpha }^3\cdot {\alpha }^6\\ & =(3{\alpha }^2-4\alpha +1)(-5{\alpha }^2-11\alpha +4)\\ & =-15{\alpha }^4-13{\alpha }^3+51{\alpha }^2-27\alpha +4\\ & =-15(5{\alpha }^2-11\alpha +3)-13(3{\alpha }^2-4\alpha +1)+51{\alpha }^2-27\alpha +4\\ & =-63{\alpha }^2+190\alpha -54.\end{array}$$Then $$\begin{array}{rl}{\alpha }^9+p{\alpha }^6+q{\alpha }^3+r& =(-63{\alpha }^2+190\alpha -54)+p(-5{\alpha }^2-11\alpha +4)+q(3{\alpha }^2-4\alpha +1)+r\\ & =(-5p+3q-63){\alpha }^2+(-11p-4q+190)\alpha +(4p+q+r-54).\end{array}$$We want this to reduce to 0, so we set $$\begin{array}{rl}-5p+3q& =63,\\ 11p+4q& =190,\\ 4p+q+r& =54.\end{array}$$Solving, we find $(p,q,r)=\boxed{(6,31,-1)}.$ For these values, ${\alpha }^9+p{\alpha }^6+q{\alpha }^3+r$ reduces to 0 for any root $\alpha$ of $x^3-3x^2+4x-1,$ so $x^9+p{x}^6+q{x}^3+r$ will be divisible by $x^3-3x^2+4x-1.$