| a | + | b | + | c | = 0 |
| z | x | y |
(1).
Multiply by z, x and y, and divide the sum by xyz; then
| a | + | b | + | c | = 0 |
| y | z | x |
(2).
From (1) and (2) by cross multiplication we obtain
| 1 | = | 1 | = | 1 | = | 1 | (suppose) |
| y (b² − ac) | z (c² − ab) | x (a² − bc) | λ |
(3).
Substituting for x, y and z in x (x − a) = yz we obtain
| 1 | = | 3abc − (a³ + b³ + c³) | ; |
| λ | (a² − bc) (b² − ac) (c² − ab) |