410 
PROFESSOR CAYLEY ON THE PROBLEM OF 
and then 
S=(X— 2)(#— 3)(X— 3), l' = (x — 2)(X — 3)(# — 3), 
whence 
d=(<r-3)(X-3)(X+ff-4) 
+ (_2X-2^+18)(-2X-2^+2+l) } 
which is 
= X 2 ( x+ 1) 
+X (x 2 — 2x- 19) 
+ x 2 — 19# 
+!;( — 2X— 2#+18). 
And we then, by means of the equations 
( #-4)2r=( #— 4)(X 2 — X + 8# -3|), 
(X— 4)2S =(X— 4)( # 2 -# +8X-3|), 
( #— 3 )i =( x — 3)( — 3# -j- £), 
(X-3)* =(X— 3)( ~3X + |), 
verify that 
d=(tf_4)2r+(X-4)2S+(#-3> + (X-3)*. 
33. Correspondence («, 0 ). 
From the values of d — <$ — cf, c—y— y' we have 
e— g — s' = ( — X 2 +13X+4#— 54)A, 
and then 
g= g '=(X-2)(ar-3)(X-3Xar-3); 
that is 
e=2(#-3) 2 (X-2)(X-3) 
+ (_X 2 +13X+4#-54)(-2X-2#+2+i), 
which is 
= X 3 ( 2) 
+X 2 ( 2# 2 -10# -10) 
+X (-10#*+26# +44) 
+ 4# 2 +44# 
+ g (— X 2 +13X+4#— -54), 
and then 
(x— i)(x— 5)2 t=(x— 4)(x — 5)(X 2 — X+8#— 3|), 
{(X-4)(X-5)+#--3}2&={(X-4)(X--5)+#-3}(;r 2 -# + 8X-3i), 
{ 3(#— 3)(# — 4)+# — 3}/ =(# — 3)(3#— 11)( — 3#+£) 
2(X — 3)(X— 4)«=2(X— 3)(X— 4)( — 3X+|) ; 
and summing these values and comparing, 
c=(x—i)(x— 5)2r + 2(X— 3)(X— 4)/t 
+[(X-4)(X~5)+a?~3]2^+[3(^-3)(a:-4) + a:-3> 
