If x, y, z, p, q, r are distinct real numbers such that
$$\left(\frac{1}{x+p}+\frac{1}{y+p}+\frac{1}{z+p}\right) = \frac{1}{p}$$
$$\left(\frac{1}{x+q}+\frac{1}{y+q}+\frac{1}{z+q}\right) = \frac{1}{q}$$
$$\left(\frac{1}{x+r}+\frac{1}{y+r}+\frac{1}{z+r}\right) = \frac{1}{r}$$,
find the numerical value of $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$$.

Substituting $$t=\frac{1}{p}$$ and manipulating the equal expressions leads to a cubic equation in t whose three roots are $$\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$$. Comparing the resulting cubic's coefficient of $$t^2$$ shows it vanishes, and since the sum of roots of a cubic equals the negative ratio of the $$t^2$$ coefficient to the leading coefficient, this sum must be 0. So $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$$.