28 ALGEBRAIC INVARIANTS 
of f. To see the nature of these functions, let x\ — \X2~ 
be a factor of f. After replacing xi by \x2+uxs in f, we 
obtain a cubic function of X2 and X3 whose four coefficients 
must be zero. Eliminating X and ¡i, we obtain two conditions 
involving r and the coefficients of / rationally and integrally. 
The greatest common divisor of their left members is the 
required quartic function of r. Unless the coefficient of r 1 is 
constant, a root would be infinite for certain /’s. The inflexion 
triangles of a general cubic curve /=0 are given by h-\-24rf =0, 
where h is the Hessian of f and r is a root of the quartic (1) in 
which a and b are rational integral invariants of f. 
The explicit expressions for these invariants are very long; 
they are given in Salmon’s Higher Plane Curves, §§ 221-2, 
and were first computed by Aronhold. For their short sym 
bolic expressions, see § 65, Ex. 4. 
EXERCISES 
1. Using the above inflexion triangle yiy 2 y 3 =0, where 
rxi—x 2 =y u Vrxg ± {rx 2 +kxi) = 2y 3 , 2y 3 , 
k=(r 2 +3b)/2, r 2 +£ = f(r 2 +6)^0, 
as shown by use of (1), we have the transformation 
Vns 3 =y 2 +y 3 , (r 2 +k)xi=ryi+D, (r 2 +k)x 2 =—ky 1 +rD, 
where D — y 2 —y 3 . Using (1) to eliminate a, show that 
^ i r ‘ 2 +6} P = - (y** ■— y» 3 ) +3)Wi - 7 (r 2 +9%1*. 
of o 
Adding the product of the latter by 54 to its Hessian, we get the product 
of yiy 2 y 3 by 3 b {r 2 +b)/r 2 . Hence the nine points of inflexion are found by 
setting yi, y 2 , y 3 equal to zero in turn. 
2. By multiplying the y’s in Ex. 1 by constants, derive 
f=a(Zi 3 +Zi 3 +Z 3 3 ) +6/3ZiZ 2 Z 3 , 
called the canonical form. Its Hessian is 6 3 h, where 
h = —a/3 2 (Zi 3 +Z 2 3 +Z 3 3 ) + (a 3 -f-2/3 3 )ZiZ 2 Z 3 , 
Thus find the nine inflexion points and show that the four inflexion triangles 
are 
ZxZ 2 z 3 = 0, 2zp - S/ziZjZs = 0 (/ = 1, W, W 2 ),