110 
ON A CERTAIN SEXTIC TORSE. 
[436 
if for shortness 
U' = y-z . c 2 /i 4 / 4 (3b 2 g 2 — c l h- ) 
+ yz-. b~g i f i (3c 2 h 2 — b 2 g 2 ) 
+ z 2 x . a 2 / 4 / (8c 2 h 2 — a 2 / 2 ) 
+ 2^. c 2 /i 4 / (3a 2 / 2 — c 2 /i 2 ) 
+ x-y. 6 2 //i 4 (3a 2 / 2 — 6 2 / ) 
+ aa/ 2 . a 2 / 4 /i 4 (36 2 / — a 2 / 2 ). 
Substituting the 0-values, the terms of U, other than £T, are at once seen to 
contain the factor (0 + a) (0 + /3) (0 + 7), and we have 
M — 3abc ( a 2 f 6 (h 4 y 2 — 7h 2 g 2 yz + g i z 2 )' 
< + b 2 g 6 (f 4 z 2 — 7f 2 h 2 zx + lb 4 x 2 ) - 
l + c 2 /^ (/a; 2 - 7//Vy +/y) , 
+ 7 (a 4 / 4 + 6y + <?k*)f*fh* abc (0 + a) 3 (0 + ¡3) 2 (0 + 7)' 
+ M\ 
where 
U' + (abc) 3 (6 + 3) 4 P 3 Q = (6 + a) (0 + /3) (0 + 7) M\ 
18. Write for shortness p, q, r — (af, bg, ch); after a complicated reduction, I obtain 
3abc M' = a 2 g 2 h 2 (r — p) (p — q) (— 2p 4 + op'-gr — 6q 2 r 2 ) a? 2 
+ b 2 h 2 f 2 (p — q)(q — r)(— 2/ + 5/rp — Qq 2 p 2 ) y 2 
+ c 2 / 2 / (5 — r ) (?• — p) (— 2r 4 + 5r 2 pg — Op 2 /) 2 2 
+ 2/ 2 g 2 h 2 b 2 c 2 (7p 4 — 20p 2 qr -f 4/r 2 ) yz 
+ 2f 2 g 2 h 2 c 2 a 2 (7/ — 20pq 2 r + 4r®p 4 ) 2a? 
+ 2f 2 g 2 li 2 a 2 b 2 (7r 4 — 20pqr 2 + 4p 2 /) X V 
— 2f 2 g 2 h 2 (p i + g 4 + r 4 ) (a 2 a: + 6 2 a/ + c 2 2) 2 . 
We then have 
9a6c if = terms (a?, 2/ + 9a6c ilP, O = terms (a;, y, 2) 
as above; and 
27ikT a&c + 30 + (*) = 0, 
which gives (*). 
19. After all reductions we find: 
— x (*) = a 2 g 2 h 2 (28p 6 — 84p 4 qr + Q2p 2 q 2 r 2 — 28/r 3 ) x 2 
+ b 2 h 2 f 2 (28q 6 — 84pq*r + 62p 2 q 2 r 2 — 28r 3 p s ) y 2 
+ <?f 2 g 2 (28r 6 — 84<pqr + 62p 2 q 2 r — 28p 3 q 3 ) z 2 
4-/ 2 (— 3/ + 14>p 6 qr — 130p 4 q 2 r 2 — 136p 2 q 3 i^ + 4>2q*r*) yz 
+ g 2 (— 85 s + 14>pq B r — 130p 2 q*r 2 — V8Qp 3 q-r 3 + 42?y 4 ) zx 
+ h 2 (— 3r 8 + 14>pqv r> — 130p 2 q 2 7 A — 136p 3 q 3 r 2 + 42p l q*) xy;