94 
ON A CERTAIN SEXTIC DEVELOPABLE 
[398 
that is 
or, what is the same thing, 
xyz (3w 2 + x 2 + y 2 + z 2 ) — 2w (y 2 z 2 + z 2 x 2 + x 2 y 2 ) = 0, 
we have to show that this is in fact included in the former system, for then the 
four equations with 0 eliminated will it is clear give two equations only. 
Observe that the former system may be written 
scjjif-vz 2 ) 
(/j, - v) yz 
(/1 — v) y 2 z 2 + (v — X) z 2 af + (\ — /x) x 2 y 2 = 0, 
fJb — V 
in the equation 
xyz (3w 2 + x? 4- y 2 + z 2 ) — 2w (y 2 z 2 + z 2 x 2 -f x 2 y 2 ) = 0, 
(y — v) y 2 z 2 + (v — \) z 2 x? + (A — y) x?y 2 = 0. 
the result is 
The substitution in question gives 
Sx 2 (fiy 2 — VZ-f 
{fi-vfyz 
2 (yy 2 - vz 2 ) 
(fl-v)yz 
+ yz(a? + y 2 + z 2 ) 
(;y 2 Z 2 + Z 2 Xr + x 2 y 2 ) = 0, 
that is 
Sx 2 (yy 2 — vz 2 ) 2 4- (y — v) 2 y 2 z 2 {x 2 + y 2 + z 2 ) — 2 (y — v) (jiy 2 — vz 2 ) (y 2 z 2 4- z 2 xr 4- x 2 y 2 ) = 0, 
which is in fact 
— y 2 y 2 (y 2 — z 2 ) (z 2 — x 2 ) + 2yvx 2 (y 2 — z 2 ) 2 — v 2 z 2 (y 2 — z 2 ) (x 2 — y 2 ) = 0, 
that is, throwing out the factor y 2 — z 2 , it is 
— y 2 y 2 (z 2 — x 2 ) + 2 yvx 2 (y 2 — z 2 ) — v 2 z 2 (x* — y 2 ) = 0. 
But in virtue of the equation ^ 4- - 4- - = 0, we have 
A. y v 
^ {(/i — v) y 2 z 2 4- (v — X) z 2 x 2 4- (\ — y) x 2 y 2 ) 
A- 
= (y- - z 2 ) + yy 2 (z 2 -X 2 ) 4- vz 2 (of - y 2 )l 
= fJLV X? (y 2 — Z 2 ) — (¡l + v ) [^2 (¿2 _ gp) q. v2 y _ y 2)] > 
= — yry 2 (z 2 — X?) + 2/JbV X 2 (y 2 — z 2 ) — V 2 Z 2 (x 2 — y 2 ), 
and the required property thus holds good.