Let $P(x)=x^3+qx+r$. I have to show that the calculation of the three roots $\lambda_i(q,r),i=1,2,3$ can be extremely ill conditioned.
For this I looked at the implicit derivative of $P(\lambda_i(q,r))=0$. I get $\dfrac{\partial\lambda_i}{\partial q}=-\dfrac{\lambda_i(q,r)}{3\lambda_i(q,r)^2+q}$ and $\dfrac{\partial\lambda_i}{\partial r}=-\dfrac{1}{3\lambda_i(q,r)^2+q}$ thus for the relative condition $-\dfrac{q\lambda_i(q,r)-r}{3\lambda_i(q,r)^3+q\lambda_i(q,r)}$. I am unable to show that this can get bigger than any bound. Any help is appreciated.
Consider the polynomial $x^3+5x-4$. It has a minimum at $(\sqrt {\frac 53},4-\frac {10}3\sqrt {\frac 53})\approx (1.291,-0.303315)$ bracketed by roots at $1, \frac 12(\sqrt{17}-1)\approx 1.56155$. As we increase the constant term we get close to having a double root. $\frac {d \lambda_i}{dr}$ is the inverse of the slope at the root. Since we can make the slope arbitrarily small by making the minimum arbitrarily close to the real axis, we get ill conditioning. If we solve $x^3-5x+4.3033=0$ the two positive roots are about $1.28904,1.29295$ If we change it to $x^3-5x+4.30331$, the roots move to $1.28978,1.292111$. They have moved $75 \cdot 10^{-5},84 \cdot 10^{-5}$ for a change of $1\cdot 10^{-5}$ in $r$