398 
ON THE JACOBIAN SEXTIC EQUATION. 
[776 
where Y, Z denote X + 10(7, X — 10(7 respectively, and consequently YZ = X 2 — 100(7 2 . 
Hence, writing as before a, /3, 8, </> to denote ag, bf d 2 and a 2 / 3 + b 3 g 2 respectively, the 
result finally is 
1 
0 
a + 3 
0 
a 2 + 3 
a 2 / 3 +216 
J8 
+ 
2160 
/3+72 
a/3 + 36 
b 3 g 2 — 216 
a 3 
+ 
1 
8 - 300 
aS - 600 
a 2 /3 
- 
36 
f3 2 + 1296 
a 2 S 
- 
30 
/88 — 144 00 
a/38 
- 
360 
S 2 + 30000 
aS 2 
+ 
30000 
/3 2 S 
+ 
12960 
(38 2 
+ 
720000 
8 s 
- 
1000000 
where observe that the coefficient of the term in X is 216 (a 2 / 3 — b 3 g 2 ), = 216 V($ 2 — 4a 2 /3 3 ). 
We have as before ag + 9bf— 20<7 2 = 0, that is, a+ 9/3 — 208 = 0; and using this equation 
to eliminate a, also in the constant term writing its value for (/> in terms of h, 
</> = A + (- 60/3 2 + 240/38 - 25 6 8 2 ) V8, 
the new equation is 
- 5 X 
5 x 
5 x 
1 
0 
(3- 9 
8 + 48 
0 
/8 2 - 24 3 
/38- 1872 
S 2 +3840 
-216VA 
hJ8 + 432 
/8 3 - 729 
/3 2 S + 4184 
/3S 2 - 11520 
8 s + 8292 
1) 6 = 0, 
where 
A = {h + (- 60/3 2 + 240/38 - 2 5 68 2 ) VS} 2 - 4 (- 9/3 + 20S) 2 /3 3 
= h 2 4- 2/i . 
/3 2 -60 
/88 + 240 
8 2 - 256 
4 (/3 — 48) 3 (9/3 - 168) 2 . 
It is to be shown that this Tschirnhausen-transformation of the Jacobian sextic is, 
in fact, the resolvent sextic of the quintic equation 
(a, 0, c, 0, e, f$#, l) 5 = 0, 
where 
a = 1, c = 2(7, e = — 9bf+ 36<7 2 , f 2 =216A.