394 
PKOFESSOE CAYLEY ON THE PKOBLEM OF 
and we thus obtain the three equations 
0=108 — 6L—3Z—18X, 
0= 88— 4L-3Z— 12a, 
0= 81 — 3L— 3/— lOx, 
giving L = l, Z=16, a =3. Whence, finally, 
<px= {X 2 +X(2x 3 -10x 2 +12x- l)-ix 3 +20x 2 -16x- 3|}DF. 
Second process, by correspondence. We have 
g -/,-%' + F(e— s— g')=0, 
e — s — £' + D(c— y— y')=0, 
and thence 
Moreover 
g— X— %' = DF(c— y— •/). 
X =(X-2)(x-3)D(x-l)F(x-l), 
% '=F(x-l)D(x-lXX-2)(x-l), =& 
%+x'=DF(X-2)2(x-3)(x-1A 
and 
c _ y — y ' = 2r-h(X-3>— 2(X— 2)(tf— 3), 
as is easily obtained, but see also post, No. 29 ; hence 
g=DF multiplied into 
(X— 2). 2(ir— 3)(#— l) 2 
+(X-2).-2(*-3) 
+ 2r+(X — 3 )%■ ; 
but I reject the term DF.(X — 3)z as in fact giving a heterotypic solution ; I do not go 
into the explanation of this. And then substituting for 2r its value, we have 
g=DF multiplied into 
(X—2).2x(x—l)(x—2) 
+X 2 -X+8a,*-3|, 
where the second factor is 
=X 2 + X(2x 3 - 1 Ox 2 + 12x- 1) ■ - 4tf 3 + 20x 2 - 16a:- 3f , 
which is the foregoing result. 
Case 40. B = D = F=tf=„r. By reciprocation of 39, 
No. = {^ 2 +.r(2X 3 — 1QX 2 +12X— -1) — 4X 3 -f20X 2 — 16X— 3f }«c. 
