351] 
ON CUBIC CONES AND CURVES. 
409 
On the relation of the two forms a? + y 3 + z 3 + 61 xyz = 0, and (X + Y + Z) 3 + 6kX YZ = 0. 
Nos. 21 to 24. 
21. Starting with the form x 3 + y 3 + z 3 + 6lxyz = 0, and writing 
then we have 
x = 
— 2 lx + y + z, 
Y = 
x — 2 ly + z, 
Z = 
x + y — 2 Iz, 
XYZ = 
— 21 (a? 3 + y 3 + z 3 ) 
+ (1 — 21 + 4>l 2 ) (y 2 z + yz 2 + z 2 x + zx 2 + x 2 y + xy 2 ) 
+ 2 (1 — 31 — 41 3 ) xyz, 
X+Y+Z= 2 (1 - l)(x + y + z), 
and thence 
(X + F + Z) 3 = 8(1 — I) 3 (x 3 + y 3 + z 3 ) 
+ 24(1 — I) 3 (y 2 z + yz 2 + z 2 x + zx 2 + x 2 y + xy 2 ) 
+ 48 (1 — I) 3 xyz, 
and we thus obtain 
(1-21 + U 2 ) (X+Y+Z) 3 + 24 (l -1 ) 3 XYZ 
= 8 (21 + l) 2 (l — l) 3 (x 3 + y 3 + z 3 + 6lxyz) ; 
or, what is the same thing, 
if 
(X + Y + Z) 3 + QkXYZ - 8 (2 * + 1)8 (a 3 + y 3 + z 3 + Qlxyz), 
k = 
4(Z-1) 3 
1-21 + M 2 ' 
22. For the form 
we find 
(X + Y+ Z) 3 + QkXYZ — 0, 
S= (1 +ky, 
— 6 (1 + k) 2 
+ 8 (1 + k) 
-3 
= k 3 (4 + k) 
T = - 8(1 +k) 3 
+ 72(l + ky 
- 128 (1 + k) 3 
+ 72(1 +k) 2 
- 8 
= — 8k 1 (6 + 6k + k 2 ), 
C. V. 
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