152 Proceedings of Royal Society of Ediiiburgh. [sess. 
easiest. way of obtaining derived equations is to multiply the lower 
equation by x, y, and and then to transform the three multiples 
so as to remove the 2nd and 3rd powers of Each of the trans- 
formed multiples in combination with the higher equation will then 
furnish an independent partial eliminant. Thus from the complete 
quadratic (A) here given, we form the three multiples (a'), (6'), (c'), 
and then by removing we form the expressions (a), (5), (c), as ? 
thus : — , 
a-^x^ + a. 2 xy -f a^y"^ -p a^z -f- ayyz -P a^z‘^ = 0 . . (A) p 
a^x^ - 1 - -p a^xy^ -p a^x^z -p a^xyz -p (agx)z^ - 0 . . . . 
ajX^y + a^if- -p a^y^ -p a^xyz -p a^yh. -p ( agy)?J^ = 0 . . . . 
{a) ) 
■ (^')h 
e')' 
{ («i + afjx -p -p { -p (ag -p af)xy -p agxz'\y 
{% + + [W + (% + af)if -p a^xz + ayyz}y 
II II 
o o 
• («)) 
• W' 
{a^x -p {a^ -p -p a^xy -p cf^y^ + 02xz + (a^ -P aQ)?jz}y = 0 . . (c) 
I next write the original cubic (B), in which the 2nd and 3rd 
powers of ^ are supposed to be removed by substitution, and then 
the three derived equations formed respectively by combining (B) 
and (a) ; (B) and (b) ; (B) and (c) ; putting for + ; and 
for ^3 -p ag — 
{ bjX -p b2^}x^ -p { -p b^xy -p b^if -p b^^z -P b^yz] y = 0 . . (B) 
{a-^x -P a^z){bgX^ -P b^xy -P b^y"^ -p bgxz -p b^^yz) - {a^x‘^ -P a^xy 
+ ayxz){b-^x + b2z) = i) . . ... . ( 1 ) 
{a^y){b^x'^ + b^xy -p bgy‘^ -P bgXZ + b^yz) - {a^xy -P a^y‘^ 
-\-a^xz^agijz){b^-\-bf)^^ (2) j 
(a^x -p a^z)(bgx^ -p b^xy + b^y^ + bgXz -P b^yz) - (a^x^ -p a^xy -p a^y^ 
+ a2Xz-hagyz)(bjX-{-b2z) = 0 ..... (3) 
The determinant is — 
