66 
V 3 + 3aY 2 + 3(a 2 - xy) Y - x 3 - y 3 - 2>axy + a 3 = 0 (44) 
The remaining two roots must be obtained from 
V 2 + (x + y + 2a) Y + x 2 + y 2, - xy + ax + ay + a 2 = 0 (45) 
11. The values of x, y in (44) can be found by a quadratic 
so as to make (44) coincide with the classical cubic. 
V 3 + 3aV 2 + 36V + c - 0 (46) 
For this purpose the equations 
a 2 - xy = b (47) 
x 3 + y 3 + 3 axy = a 3 -c (48) 
will be necessary. 
From these two equations there results 
. 3a& - 2a 3 — c 1 — 7 — 
x 3 = g + ^ V (3 ab - 2a 3 - cf - 4(a 2 - b) 3 
n 3 ab — 2 a 3 — c 1 
ys - y/ (3ab - 2a 3 - c) 2 - 4(a 2 - b) 3 
The solution of a quartic by a partial differential resol- 
vent with three independent variables. 
12. Let 
V 4 + tV 2 + RV + S = 0 ,....,...(49) 
be a quartic in which t, R } S are functions of the three 
variables x, y , z. 
Differentiate (49) with respect to x, y , z respectively, then 
, . TTO o tt dt _ T0 <7R TT dS . 
(4Y 2 + 2tV + R)-t- + -r-V 2 + V + = 0 
v ' ax ax ax dx 
,(50) 
( ^ +2 ,y + E)f + |.y^|.y + | = o (51) 
(4V* + JMV + E)g + Jy» + .f.V + f = 0 (52) 
From (50) and (51) ; from (51) and (52) there results 
'dt 2 dR„ d8\dV 
^ + Ty y + Jy 
dt 
d R _ cZSyV 
^ V + dy V + 
)£-( 
dV _ /dt' 
dz \dz 
dt y 2 dR y dSWY 
dx dx dx) dy‘ 
2 dR dS\dY 
V +-T’ V + w Hr* 
a , 2 dz / dy 
(53) 
(54) 
Equation (50) is the first partial differential resolvent of 
the quartic (49) together with the two conditional equa- 
tions (53) and (54). 
The value of V, in functions of x, y , 0 , which satisfies (50), 
