[346 
291 
4 
346] THE RESULTANT OF TWO BINARY CUB1CS. 
But the two equations are 
(a, b, c, d\x, yf = 0, 
x 3 + y 3 = 0, 
the last of which gives y = — x, y= — <ox, y = — (o 2 x, if &> be an imaginary cube root of 
unity, and hence the Resultant is 
= (a — 3b + 3c — d) (a — 3bw + 3ca> 2 — d) (a — 3box + 3cw — d), 
which is 
= (6 — 3b + 3c) (6 — 3bw + 3cco 2 ) (6 — 3bco 2 + 3Co)), 
or finally is 
= 6 5 - 27b 3 + 27c 3 + 27bc6, 
and the formula is thus verified. 
If the two cubics are taken to be 
(a, b, c, d\x, y) 3 = 0, 
(b, c, d, e^x, y) 3 = 0, 
then the formula gives for the Discriminant of the quartic function {a, b, c, d, e§x, y) 4 
a new expression, which however does not appear to be an elegant one. 
37—2