402 
PROFESSOR CAYLEY ON THE PROBLEM OF 
+xx' . 4X 3 +12 X 2 X' + 1 2XX' 2 + 4X' 3 
— 2 SX 2 — 5 6XX' — 2 8X' 2 
+ 56X+56X' 
-22 
+a ,s . 2X 3 + 6X 2 X' + 6 XX' 2 
— 14X 2 — 28XX' 
+28X 
+x. -30X 2 X'-30XX' 2 -1GX' 3 
+ 140XX'-f 70X' 2 
— 116X'— 6|' 
+at. — 10X 3 — 30X 2 X'— 30XX' 2 
+ 70X 2 -[-140XX' 
-116X-6I 
+ 36X 2 X'+36XX' 2 
— 152XX' 
-4 (Xi'+X'i); 
whence 
<px—a multiplied into 
A + 1 ) 
+r>( 2X 3 -14X 2 + 28X-11) 
+^(-10X 3 +70X 2 -116X +0 
+ 12X 3 — 76X 2 +LX 
+£ ( — 6^— 4X +^), 
where the constants l, L, A have to be determined. We should have <px=0 for a cubic 
curve; viz. #=3: X=6, £ = 18; X=4, £ = 12; or X=3, £=10. Writing first x— 3, 
the equation is 
8X 2 -96X-72-£(18+4X)+3Z + XL + £a=0, 
giving in the three cases respectively 
3Z+6L+18A=1116, 
3£+4L+12x= 736, 
3Z + 3L+10X= 588; 
and we have then /= — 8, L=64, X=42, so that the required number is 
