776] 
ON THE JACOBIAN SEXTIC EQUATION. 
399 
I consider the general quintic (a, 5, c, d, e, f^x, 1) 5 = 0; taking the roots to be 
Xi, x 2 , x 3 , x 4 , x 5 , and writing 
04 = 12345 - 24135, 
0 2 = 13425 - 32145, 
0 3 = 14235 — 43125, 
0 4 = 21435 - 13245, 
0 5 = 31245 - 14325, 
0 6 = 41325 -12435, 
where 12345 is used to denote the function 
= (xjx 2 + x& 3 + x 3 x 4 + XiX 5 + x 5 Xj) V(20), 
(this numerical factor \/(20) being inserted for greater convenience), then the equation 
whose roots are 0 1} 0 2 , 0 3 , 0 4 , 0 5 , 0 8 , which equation may be regarded as the resolvent 
sextic of the given quintic equation, is 
a 6 x 
— 5a 4 x 
5a 2 x 
a 2 
+ 5 
1 
' 0 ' 
ae 
2 a?df 
+ 1 
+ 1 ct 3 cf 3 
— ibd 
+ 3 a 2 e 2 
— 2a?def 
+ 3c 2 
&c. 
+ &c. 
1) 6 = 0, 
□ =a 4 / 4 + &c., the discriminant of the quintic: see p. 274* of my paper “On a new 
auxiliary equation in the theory of equations of the fifth order,” Phil. Trans, t. CLI. 
(1861), pp. 263—276, [268]. 
I now write 5 = 0, d = 0, but, to avoid confusion again, write roman instead of 
italic letters, viz. I consider the resolvent sextic of the quintic equation 
(a, 0, c, 0, e, l) 5 . 
Many of the terms thus vanish, and the equation assumes the form 
a 6 x 
— 5a 4 
5a 2 
| 1 
1 □ 
+ 5 
' 1 ' 
0 
ae + 1 
a 2 e 2 + 3 
+ 1 
a 3 cf 2 + 1' 
c 2 + 3 
ac 2 e — 2 
a 3 e 3 + 1 
c 4 + 15 
a 2 c 2 e 2 - 11 
ac 4 e +35 
c 6 - 25 
and then if, as before, 
a = 1, c = 2d, e = — 95/+ 36c£ 2 , f 2 = 216/i, 
or say 
a = 1, c = 2 \/8, e = — 9/3 + 36S, f 2 =216/i, 
* [This Collection, vol. iv., p. 321.]