jmc

From the given equation, $$\begin{array}{rl}a& =k(1-b),\\ b& =k(1-c),\\ c& =k(1-a).\end{array}$$Then $$\begin{array}{rl}a& =k(1-b)\\ & =k(1-k(1-c))\\ & =k(1-k(1-k(1-a))).\end{array}$$Expanding, we get $a{k}^3+a-k^3+k^2-k=0,$ which factors as $$(k^2-k+1)(ak+a-k)=0.$$If $ak+a-k=0,$ then $a=\frac{k}{k+1},$ in which case $b=c=\frac{k}{k+1}.$ This is not allowed, as $a,$ $b,$ and $c$ are distinct, so $k^2-k+1=0.$ The sum of the roots is $\boxed{1}.$

Note: The roots of $k^2-k+1=0$ are $$\frac{1\pm i\sqrt{3}}{2}.$$For either value of $k,$ we can take $a=0,$ $b=1,$ and $c=k.$