576 PEOEESSOE CAYLEY ON THE SEXTACTIC POINTS OE A PLANE CUEVE. 
and thence multiplying by B 1? =5, and with the result operating upon U, we find 
3^u= ( ^l ( 7- 2) ^3u-^a^vu+^sv»u. 
But BU=0, and thence also V(BU)=0, that is (V .B)U + VBU = 0; moreover V . B=0, 
and therefore (V. B)U=0, whence also VBU=0. Therefore 
or substituting for BV 2 U its value =<I>BH — HB<1>, we have the required expression for 
B^U. 
70. Proof of equation B 2 B 3 U = j~_ (( 3 m — 6)HB d> -f ( — m-f- 3)<1>BH) + ^ 3 - { — (B . V)H[ . 
We have 
+ V, 
and thence multiplying by B?=B 2 , and operating on U, 
B^U^-^BWU- -J- $B 3 U + -^(B . V)B 2 U. 
i 3 m— 1 m— 1 1 in — l v ' 
To reduce (B . V)B 2 U, we have 
and since 
B(VB 2 U)=VB 3 U+ (B . VB 2 )U 
= VB 3 U + [(B . V)B 2 + V(B . B 2 )]U 
=VB 3 U + (B . V)B 2 U+2VBB 2 U, 
multiplying by VB, and with the result operating on U, we obtain 
VBB 2 U= - ■ <E> VBU+ ——7 V 2 BU ; 
2 m — 1 1 m — 1 
or since VBU=0, this is 
VBB 2 U = -^-,V 2 BU. 
Hence m ~ 1 
B(VB 2 U)=VB 3 U+(B . V)B 2 U+~ V 2 BU, 
that is m 
(B . V)B 2 U=B(VB 2 U)-VB 3 U-~ V 2 BU. 
Substituting this value of (B . V)B 2 U, we find 
B 2 B 3 U= -^Bd>B 2 U- --7 $B 3 U 
m — 1 m — 1 
(3(V3 ! U)— V3*U) 
+p^ )2 (-2V"3U), 
the three lines whereof are to be separately further reduced. 
