556 
PROBLEMS AND SOLUTIONS. 
[485 
Solution by the Proposer. 
The equation may be transformed into the form 
(— 8x + y + zf -f (x — 8y + zf + (x + y - 8z) 3 = 0, 
and it thence follows immediately that the stationary tangents are the lines 
— 8x + y + z = 0, x — 8y + z — 0, x + y— 8z =0, 
respectively, and that the three points of contact, or inflexions, are the intersections of 
these lines with the line x + y + z = 0. 
In fact, writing 
X — kx + y + z, Y = x + ky +z, Z = x + y + kz, 
we have identically 
(X+Y + Z) 3 -27XYZ 
= (k + 2) 3 (x + y + z) 3 — 27 (kx + y + z) (x + ky + z) (x + y + kz), 
= (a: 3 + y 3 + z 3 ) {(& + 2) 3 — 27&} 
+ 3 (yz 3 + y 3 z 4- zx 2 -\- z 2 x + xy 1 + x?y) {(A; + 2) 3 — 9 (k 2 + k + 1)} 
+ 3 xyz {2 (k + 2) 3 — 9 (k 3 + Sk + 2)} 
= (k— l) 2 (k+8)(x? + y 3 + z 3 )+3(k — l) 3 (yz 2 + y 3 z -\-zx 2 -\-z 2 x -\-xy 2 + x 2 y)—S(k— 1) 2 (7k + 2)xyz. 
Hence, writing k = — 8, we have 
(X + F+ Z'Y — 27XYZ = — 2187 {yz 2 + y 2 z + zx 2 + z 2 x + xy 2 + x 2 y — 6xyz], 
= — 2187 [x (y — z) 2 + y (z — x) 2 + z(x — y) 2 }. 
The equation of the given curve is therefore 
(X +Y+ Z) 3 — 27XYZ = 0, or X h + Y k + Z* = 0, 
where of course X, Y, Z have the values 
X = — 8 x + y + z, Y = x — 8y + z, Z = x + y — 8z. 
[Vol. vi. pp. 35—39.] 
1990. (Proposed by Professor Sylvester.)—Prove that the three points in which 
a circular cubic is cut by any transversal are the foci of a Cartesian oval passing 
through the four foci of the cubic. 
Solution by Professor Caylev. 
Some preliminary explanations are required in regard to this remarkable theorem. 
1. I call to mind that a circular cubic (or cubic through the two circular points 
at infinity) has 16 foci, which lie 4 together on 4 different circles; and that the