148 
The selection, as it appears to me, must lie between the 
above quartic and the quartic (13). 
7. Differentiate (2) with respect to x, where a, b, c, m 
dv 
are functions of x, and put as usual = y\ &c. 
. 1 o}y 3 + 3&y + 3 dy + m 1 _ 
^ + 3 ay 2, + 6by + 3c 
Eliminate y s by means of (2) 
(< ah 1 - bad)y 2 + (ad - ca})y + ^(am 1 - ma l ) 
a?y 2 + 2aby + ac 
Put (ab 1 - ba l )y 2 + (ac 1 - cad)y + \(am l - ma 1 ) 
= 0 (18) 
= ( a*y 2 + 2 aby + ac)(Py 2 + Qy + R) (19) 
Then (18) becomes 
y 1 + Vy* + Qy + R = 0 (20) 
which is the first resolvent of the cubic (2). 
Multiplying the right-hand side of (19), and, eliminate 
y\ y s by means of (2), there results 
(3 6 2 - 2ac)P - a6Q + a 2 R = ab 1 - ba 1 
(36c- am ) P - 2acQ + 2a6R = ac 1 - ca 1 
6mP - amQ + acR = ^ (am 1 - ma 1 ) (21) 
The values, therefore, of P, Q, R, which satisfies equation 
(20) are to be determined from the system (21), and are as 
follows 
Put a — 3(36 2 c 2 - 4ac 8 - 46 3 m + 6abcm - a 2 m 2 ) (22) 
aP = (46 2 m - 36c 2 - acm)a 1 
4- 6a(c 2 - bm)b x 
+ 3 a(am - bc)d 
+ 2a(6 2 - ac)m l (23) 
aQ = (76cm - 6c 3 - am^a 1 
+ 3 (36c 2 - 26 2 m — acm)b l 
+ 3(2ac 2 + abm - 36 2 c)c 1 
+ (66 3 + a 2 m- 7abc)m x (24) 
aR = 2m(6m - c 2 ) a 1 
+ 3m(6c - am)6 J 
+ 6m(ac - 6 2 )c 1 
4 (36 2 c + abm - 4ac 2 )m x 
( 25 ) 
