146] A MEMOIR ON CURVES OF THE THIRD ORDER. 415 
which shows that the point in question lies on the cubic. We have thus the 
theorem : 
The first or conic polar of a point of the cubic touches the cubic at the point, 
and meets it besides in four points, which are the angles of a quadrangle the 
centres of which lie on the cubic. In other words, the quadrangle is an inscribed 
quadrangle. 
40. To find the equations of the three axes of the quadrangle, that is of the 
lines through two centres. 
We have 
(4/c 2 YZ - A 2 X 2 ) x + ( A 2 XF + 2k\Z 2 ) y + ( A 2 XX + 2rc\Y 2 ) z = 0, 
( A 2 XF + 2k\Z 2 ) x + (4k~ZX — A 2 F 2 ) y + ( A 2 FX + 2k\X 2 ) z = 0, 
( A 2 XX + 2k\Y 2 ) x + ( A 2 FX + 2kAX 2 )y + (4k 2 IF- A 2 X 2 ) z = 0 ; 
or arranging these equations in the proper form and eliminating k 2 , «A, A 2 , we find 
YZx, Z 2 y + Y 2 z, X (— Xx + Yy + Zz) j = 0 ; 
ZXy, X 2 z + Z 2 x , Y ( Xx — Yy + Zz) 
XYz, Y 2 x + X 2 y, Z{ Xx + Yy + Zz) 
or, multiplying out, 
XYZ {(Z 3 - F 3 )X s + (X 3 - X 3 )y 3 + (Y 3 — X 3 )£ 3 } 
+ x 2 yZY 2 (- 2X 3 +Y 3 + Z 3 ) + zx 2 YZ 2 (2X 3 - F* - Z 3 ) 
+ y 2 zXZ 2 (— 2 F 3 + Z 3 + X 3 ) + xy 2 ZX 2 (2 F 3 - X 5 - X 3 ) 
+ z 2 xYX 2 (- 2X 3 + X 3 + F 3 ) + yz 2 XY 2 (2X 3 - X 3 - F 3 ) = 0. 
We may simplify this result by means of the equation X 3 + F 3 4- X 3 + 61XYZ= 0, so as 
to make the left-hand side divide out by X YZ: we thus obtain 
(X 3 — F 3 ) x 3 + (X 3 — X 3 ) y 3 + ( F 3 — X 3 ) z 3 
+ (- 3X 2 F - 61Y 2 Z) xhy + (- 3 F 2 X - 6£X 2 X) y 2 ^ + (- 3X 2 X - 61X 2 Y) z 2 x 
+ ( 3XY 2 + 6lX 2 Z)xy 2 + ( 3 YZ 2 + 61F 2 X) yz 2 + ( 3XX 2 + 61Z 2 Y) zx 2 = 0; 
or in a different form, 
(y 3 — ^ 3 ) X 3 + (V* — ¿c 3 ) F 3 + (ic 3 — y 3 ) X 3 
+ (— 3^? 2 y — 6lz 2 x) X 2 F+ (— 3y 2 ^ — 6lx 2 y) Y 2 Z + (— 3z 2 x — 6ly 2 z) X 2 X 
+ ( 3xy 2 + 61 yz 2 ) XF 2 + ( 3yz 2 + 6lzx?) YZ 2 + ( 3zx 2 + 6lxy 2 ) XX 2 = 0, 
as the equation of the three axes of the quadrangle.