Given a = x/(y - z), b = y/(z - x), and c = z/(x - y), where x, y, and z are not all zero, then the value of ab + bc + ca is

$\begin{array}{l}\mathrm{a}=\mathrm{x}/(\mathrm{y}-\mathrm{z})\Rightarrow \mathrm{x}-\mathrm{ay}+\mathrm{az}=0 \left(1\right)\\ \mathrm{b}=\mathrm{y}/(\mathrm{z}-\mathrm{x})\Rightarrow \mathrm{bx}+\mathrm{y}-\mathrm{bz}=0 \left(2\right)\\ \mathrm{c}=\mathrm{z}/(\mathrm{x}-\mathrm{y})\Rightarrow -\mathrm{cx}+\mathrm{cy}+\mathrm{z}=0 \left(3\right)\end{array}$

Since x, y, z are not all zero, the above system has a non-trivial solution. So,

$\mathrm{\Delta }=\left|\begin{array}{ccc}1& -\mathrm{a}& \mathrm{a}\\ \mathrm{b}& 1& -\mathrm{b}\\ -\mathrm{c}& \mathrm{c}& 1\end{array}\right|=0$

or    1 + ab + bc + ca = 0