436] 
ON A CERTAIN SEXTIC TORSE. 
109 
if for shortness 
A = ayh 4 (by + c 2 h? ), 
B = b 2 h 4 f 4 (c-h? + o?f~), 
G = c 2 /y (a 2 / 2 + by ), 
M =f 4 h 2 (a 2 f-c-Ii- - 7a-f n -by + 3bhftfb? ), N = /ty 2 (- 7a?fW + o?f‘ 2 by + 3by-tfh 2 ), 
0 — g 4 / 2 (¥g-a 2 / 2 — Iby&h? + 3c 2 A 2 a 2 / 2 ), P = y 4 A 2 (- lb 2 g 2 d 2 f 2 + byph? + 3c 2 h 2 a?J 2 ), 
Q =ky (c 2 h?by — 7c 2 h?a 2 b 2 + 3a‘ 2 f‘ 2 by), R = /¿ 4 / 2 (— 7c 2 h 2 by + <S 2 h?d 2 f' 2 4- 3a 2 f' 2 by); 
and I represent the foregoing equation by 
<t> = (a 2 x + b‘ 2 y + c 2 z) 2 U + xyzPL. 
Hence, writing for x, y, z the foregoing values, we have 
= 9a-b 2 d 2 (0 + a) 2 (0 + /3) 2 (0 + y) 2 JJ + abc (0 + a) 3 (0 + /3) 3 (0 + 7) 3 H ; 
and thence 
27 U+ ^ (0 + a) (0 + /3) (0 + 7) (3.ft + (*)) = - 27 (a&c) 3 (0 + 8) 4 P 3 Q 5 
that is 
27 [ E7 + (obey (0 + 8) 4 P S Q] + ^¿(0 + <x)(0 + ¡3) (0 + 7) (3il + (*)) = 0. 
In order that this may be the case, it is clear that we must have 
U + (abc) 3 (0 + 8) 4 P 3 Q = (0 + a)(0 + ¡3) (0 + 7) M, 
viz. the left-hand side expressed as a function of 0 must be divisible by the product 
(0 + a.) (0 + /3) (0 + y). Assuming for a moment that this is so, the quotient M will be 
a function (0, 1)'* expressible in a unique manner in the form (x, y, z) 2 , and assuming 
it to be so expressed, we have 
27Mabc + 3fi + (*) = 0 ; 
which equation, without any further substitution ol the d-values of (x, y, z), gives (*) 
in its proper form as a function of (x, y, z). 
Reduction of the Equation, U + (abc) 3 (0 + 8) 4 P 3 Q — (0 + a) (0 + (3)(0 + 7) M. 
17. We have by an easy transformation 
U = (d 2 x + b 2 y + c 2 z) ( a 2 / 6 (h 4 y 2 - 7hPfyz + g 4 z 2 ) j 
■ ; + by (f 4 z 2 - 7 f 2 h 2 zx 4- h 4 x 2 ) j- 
l+c 2 /i 6 (g 4 A - +/Y)) 
+ 7 (a 4 f 4 + by + &h 4 )fyh: 2 xyz 
+ U\