PROFESSOR CAYLEY ON THE SEXTACTIC POINTS OE A PLANE CURYE. 573 
63. This leads to the expression for d 2 U ; we have 
~{ m— l) 2 
-(^LTfp 3>V(ffBJ+yB 9 +zB,) 
_l_ — yz . 
' (m — l) sV ’ 
and operating herewith on U, we tind 
b*u= 7 (ot ili &V 
2 [in- 1) 2 
2(to — 1)S 
(to — l ) 2 
ovu 
+(^rrp V!U ; 
this is 
VU=— H, V 2 U=HcJ>, 
(to— l) 2 ~(to— 1) 
64. We have B,B 2 U=0, and thence 
that is 
(B^ 2 +B 1 b 3 +B 2 )U=0, 
B 1 B 3 U=-B^ 2 U-5 2 2 U; 
or substituting the values of B 2 d 2 U and d 2 U, we find the value of d,d 3 U as given in the 
Table. And then from the equation 
(^+6B 2 d 2 +4dA+3BI+d 4 )U, 
or 
d 4 U= _(BJ + 6dft 2 + 4d,d 3 + 3B 2 )U, 
we find the value of d 4 U, and the proof of the expressions in the Table is thus com- 
pleted. 
65. Proof of equation V.d = 0. 
We have 
v. b=v. ((b*-<»b, +(Cx-a*)b, +(a^-bx)b.) 
=V . (A(pB„-iB,)+B(.B # -XB„) +CQb,-ttij) 
= VA (pB.-iB,) + VB(jB,-xB,)+ VC( xB,-a.B,) ; 
and then 
VA=(g, ...£*■> (a, vja, h, g )=HX, 
VB = (& ...IX, p, Oft B,/)=Hp, 
VC=(a --.B, c)= H„; 
or substituting these values, we have the equation in question. 
