414 A MEMOIR ON CURVES OF THE THIRD ORDER. [146 
or, if we please, considering k, X as arbitrary parameters, then the four points lie in 
the conic 
(2kX, 2kY, 2kZ, — XX, — XY, — XZ^ac, y, zf — 0, 
or, what is the same thing, they are the points of intersection of the two conics 
Xx 2 + Yif + Zz 2 = 0, 
Xyz + Yzx + Zxy = 0. 
39. Considering the four points as the angles of a quadrangle, it may be shown 
that the three centres of the quadrangle lie on the cubic. To effect this, assume 
that the conic 
(2kX, 2kY, 2tcZ, — XX, — XY, — XZf[x, y, zf — 0 
represents a pair of lines; these lines will intersect in a point, which is one of the 
three centres in question. And taking x, y, z as the coordinates of this point, we 
have 
x 2 : y 2 : z 2 : yz : zx : xy — 4/e 2 YZ — X 2 X 2 
: 4/e 2 ZX — X 2 Y 2 
: 4/e 2 XY— \ 2 Z 2 
: X 2 YZ + 2k\X 2 
: \ 2 ZX + 2/c\Y 2 
: \ 2 XY+2k\Z 2 ; 
and we may, if we please, use these equations to find the relation between k, X. 
Thus in the identical equation x 2 . y 2 — {xy) 2 = 0, substituting for x 2 , xy, y 2 their values, 
and throwing out the factor Z, we find (4/e 3 — X s ) X YZ — kX 2 (X 2 + F 3 + Zr) = 0, and 
thence, in virtue of the equation X 3 + Y 3 + Z 3 + 61X YZ = 0, we obtain 
4/e 3 + 6l/cX 2 — X 3 = 0. 
But the preceding system gives conversely, 
X 2 : Y 2 : Z 2 : YZ : ZX : XY=№yz- X 2 x* 
: 4 k 2 zx — X 2 y 2 
: 4 K 2 xy — X 2 z 2 
: X 2 yz + 2 kXx 2 
: X 2 zx + 2 xXy 2 
: X 2 xy + 2 kXz 2 . 
Hence from the identical relation X 2 .Y 2 — (XY) 2 = 0, substituting for X 2 , XY, Y 2 
their values, and throwing out the factor z, we find (4/e 3 — X 3 ) xyz — kX 2 (x s + y 3 -f z 3 ) = 0, 
and thence, in virtue of the equation 4/e 3 — X s = — 6l/cX 2 , we obtain 
x 3 + y 3 + z 3 + Qlxyz = 0,