520 PEOEESSOE CAYLEY’S EIGHTH MEMOIE OX QU AN TICS. 
No. 15 = (—ac, —Sac, —Sac, ab—ac—bc , —3 be, — 3 be, —bc'Jx, y) a 
= bcy 3 z 3 + c®V + abx 3 y 3 . 
No. 16=(0, — abc , — abc , 0 \x, yf—abexyz. 
No. 17= ( 
a 2 b — ac 2 -\- be 2 — a 2 c 
■ — 2abe)x b 
+( 
— 5 ac 2 -\- 5 be 2 
— 5 dbc)x A y 
+ ( 
-10«c 2 +106c 2 
— 2abc)x 3 y 2 
+ ( 
— 10«c 2 +10^c‘ 2 
-j-2 abc)x 2 y 3 
+( 
— 5ac 2 -\- 5 be 2 
- \-5abc)xy 4 
+ (- 
- ab 2 — ac 2 + bc 2 -\-b 2 c 
+2 abc)y b 
=(b— c)a 2 x 5 +(c-a)by+(a- b)c 2 z 5 
— abc(y — z)(z — ■ x)(x —y )(yz + zx-\- xy). 
Article No. 256 . — Expression of the 18 -tine Invariant in terms of the roots. 
256. It was remarked by Dr. Salmon, that for a quintic ( a , b, c , d, e,ffx, yf which 
is linearly transformable into the form [a, 0, c, 0, e, Of^x, y) b , the invariant I is =0. 
Now putting for convenience y=l, and considering for a moment the equation 
x(x— {3)(x— y)(x— &)(#— 0 = 0 , 
then writing: herein — for x, the transformed equation is 
where 
x(x— (3')(x— y')(x— $)(x— s')=0, 
n/3 
1 — m( 3 ’ 
ny 
1 — my' 
&c. ; 
hence m may be so determined that fl'-j-y' may be =0; viz. this will be the case if 
j3+y=2m/3y, or In order that ei'-t-s' may be =0, we must of course have 
m=^y^, and hence the condition that simultaneously j3'-\-y'=0 and g'= 0 is 
that is, ((3-\-y)h&—f3y(h-{-s) = 0. Or putting#— a for x and /3 — a, y — a, See. 
for /3, y, &c., we have the equation 
(x—u)(x—(3)(x—y)(x — 'b)(x—e) = 0, 
which is by the transformation x— a into— -r ~ — v-— changed into 
(x—cc')(x — ft')(x — y')(x—h')(x—z') = 0 
(where a'=a), and the condition in order that in the new equation it may be possible 
to have simultaneously /3' +y' — 2a'=0, fc'+g' — 2a' = 0, is 
(/3 + y — 2a)(h — a)(s — «)— (&+g — 2a)(/3— a)(y— a) = 0, 
