400] 
GIVEN PENCIL OF SIX LINES. 
107 
we have g — 0, i = 0, and therefore C' = 0 ; the equation of the pencil of tangents is 
A' 2 D' 2 + 4P' 3 P = 0, or throwing out the constant factor U, and then replacing A', B', D' 
by their values, the equation of the pencil of tangents is 
c[(a, h, k, b^x, y) 3 ] 2 + 4 [(j, l,f\x, y) 2 ] 3 = 0, 
which is the before-mentioned result. 
The coefficients a \J (c), h V (c), k V (c), b \J (c), j, l, f or (as we may call them) the 
coefficients of the cubic curve, are, it has been seen, functions of the coefficients of 
the given sextic (*$#, y) 6 ; hence the invariants S and T of the cubic curve are also 
functions of the coefficients of the sextic, and it is easy to see that they are in fact 
invariants (not however rational invariants) of the sextic. To verify this, it is only 
necessary to show that the invariants S and T are functions of the invariants of the 
functions V (c) . (a, h, k, b\x, y) 3 and (j, l, f^x, y) 2 ; for if this be so, they will be 
invariants of the function 
[c (a, h, k, b^x, y) 3 ] 2 + 4 [( j, l, f\x, y) 2 ] 3 , 
that is of the sextic. We have in fact the general theorem, that if P, Q, B,... be 
any quantics in (x, y, ...), and </>(P, Q, R, ...) a function of these quantics, homogeneous 
in regard to (x, y, ...), then any function of the coefficients of <£, which is an invariant 
of the quantics P, Q, R, ... is also an invariant of (/>. 
Considering for greater convenience the function 
(a, h, k, b\x, y) 3 
in place of V (c). (a, h, k, b\x, y) 3 , the invariants of the two functions (a, h, k, b^x, y) 3 
and (j, l, f\x, y) 2 are as follows: 
□ = a}b- — Gabhk + 4oP 3 + 46A 3 — 3 h 2 k 2 , 
V= fj-l\ 
© = j (bh — k 2 ) + l (hk — ab) + / (ak — h 2 ), 
R = + 1 a 2 / 3 
+ 6 abflj 
— 6 ahfH 
— 6 akf 2 j 
+ 12 akfl 2 
+ 1 b 2 j 3 
— 6 bhfj 2 
+ 12 bhjl 2 
— 6 bkj 2 l 
+ 9 h 2 f 2 j 
—18 hkfjl 
+ 9 k 2 fj 2 
— 8 abl 3 , 
14—2