485] 
PROBLEMS AND SOLUTIONS. 
551 
Solution by Professor Cayley. 
Write (a, b, c, cl, e§x, l) 4 = a (x — a)(x — /3)(x — 7) (x— 8); then putting for a moment 
/3 + r y+8=p, /3y + /3S + 78 = q, /3<y8 = r, and forming the equation 
(ß + ry - 28) (ß + 8 - 2y) (7 + 8 - 2/3) = 0, 
this is easily reduced to 
— 2p 3 + 9pq — 27r = 0. 
But we have 
a (oc 3 —px 1 + qx — r) (x — a) = (a, b, c, d, e\x, l) 4 , 
and hence 
6c 46 4 d 6c 46 
_L n 1 tv* — n 
<1 = —I a + a 2 , r — a a 2 — a 3 . 
1 a a a a a 
Substituting these values of p, q, r, the foregoing equation becomes, after all reductions, 
(20 a 3 , 20 a 2 6, - 16 a6 2 + 36 a% 128 6 3 - 216 abc + 108 a?d\a, l) 3 = 0, 
and from this and the equation (a, 6, c, d, e§a, 1) 4 =0, eliminating a, we should find 
the condition for three roots in arithmetical progression. But it appears from the theory 
of invariants that the result of the elimination may be obtained by writing 6 = 0, and 
expressing the result so obtained in terms of a, H, /, J. Hence, writing in the two 
equations 6=0, the first equation contains the factor 4a 2 , and throwing this out, the 
equations become 
5aa 4 4- 27ca + 27d = 0, aa 4 + 6ca 2 + 4<7a + e = 0 ; 
or multiplying the first by a and reducing by means of the second, the two equations 
become 
baa 3 + 27 ca + 27 d = 0, 3ca 2 — 7 da + oe = 0. 
The result is of the degree 5 in the coefficients, but in order to avoid fractions in 
the final result it is proper to multiply it by a 4 ; it then becomes 
625 a 6 e 3 — 4050 a 5 c 2 e 2 + 6561 a 4 c 4 e — 1890 (feed? + 13122 a 4 c 3 d 2 + 9261 al'd? = 0. 
But writing as above 6 = 0, we have 
a = a, 
c = — 
a 
H 
I 3 H 3 
a a? a 4 
J HI 4 H 3 
r + IT rr -