1892-93.] Lord McLaren on Three Ternary Equations. 265 
identical, and the 3rd and 4th factors are also identical, we need 
only multiply together the 1st and 3rd factors, and write 
+ .... (4) 
On multiplying out, dividing by and arranging the terms in 
symmetric functions, we have 
In order to express the symmetric functions in terms of the 
coefficients of (1) and (2), we observe that 
a + — = - 2 
6 +^ = €2 
0 
whence, substituting in (5), 
^.e., 
or 
4 = 0, 
, (C'2 A'2 B'2 ^A'B'C' A 
^Ub'^BU'^CA ‘ABC h 
= 0 
CC'2 + AA'2 + BB'2 - 2A'B'C' - ABC = 0 
■ • ( 6 ) 
• • ( 7 ) 
The reason why, in this solution, the eliminant is obtained in its 
true dimension (not squared), is because in the operation (4) one of 
each of the pairs of identical factors is thrown out. 
By making (1), (2), and (3) homogeneous we see that the solution 
consists in reducing the given equations to the form. 
- e-^xy + = 0 
= 0 
the eliminant of which is the quadratic, 
+ €2^ + 632 - €i€2e3 = 4 . 
