60 
Eliminate from this equation y 4 and y s by means of the 
cubic (1), then 
- a}y - m 1 = 3P( - ay* — my) + 3Q( -ay-m) 
+ (3R + aV)y 2 + aQy + aR 
or, 
(3R - ( 2aV)y 2 + (a 1 - 3mP - 2aQ)y + m 1 + aR - 3mQ = 0 
To satisfy this equation it is necessary and sufficient that 
3R - 2aP = 0 (3) 
a} - 3mP - 2aQ = 0 (4) 
m 1 + aR - 3mQ = 0 (5) 
From these equations the values of P, Q, R are readily 
found to be 
P(4a 3 + 27m 2 ) = 9 ma 1 - 6am 1 
Q(4a 3 + 27m 2 ) = 2a 2 a x + 9mm 1 
R(4a 3 + 27m 2 ) = bama 1 - ka*m l 
3R = 2aP 
Substitute these values in (2), then 
1 9 ma 1 - bam 1 a _ 2 a*a l + 9mm 1 bama 1 - 4 aW 
y = 4a 3 + 27m 2 ' y + ~WVfhrf ' y + 4a s + 27m 2 ' ' 
Equation (6) is, therefore, the first differential resolvent of 
the cubic (1). 
Hence the cubic (1) is the integral of (6), and the value 
of y which satisfies (6) is, then, a root of the cubic (1). 
2. To find the second differential resolvent of the cubic 
(1) it will be convenient to write (6) as follows : — 
y x +py* - Qy + -g - = 0 (7) 
where, (4a 3 + 27 m*)p = bam 1 - Qma 1 
Differentiate (7) with respect to x, then 
y 11 + Zpyy 1 + pY - Q y 1 - Qty + — ^ + ^-'= 0 (8) 
From (7) there results 
1 A 2 3 2apy 
yy-Qy-py — 3 — 
= Qy 2 + + mp 
