494 G. H. KNIBBS AND F. W. BARFORD. 
Since the first term is essentially positive, it follows that 
A,A;, — A; must be negative; that is, A; > A,As. 
The condition above propounded may be written as 
follows :— 
42 A,(A, A, — 43) + A,(A,4;— A,A,)} “4+ 44,4, Ae (52) 
must be negative. 
Writing 4 for A,A, — A} and p» for A.A; - A,A, the above 
condition becomes finally that : 
A,{Ayp? + 4A,\p + 44,47} must be negative. 
This is the ‘‘irreducible case’’ in Oardan’s solution of 
the cubic. 
When the equation contains the constant a we must have 
five points given on the curve and the computation becomes 
more tedious. We then proceed as follows:— 
Adopting the same notation we have for the general 
expression 
Ing = & + Ga 4 AO sien (53) 
Hence, writing L' for L (4-1), and M’ for M (6-1), the 
general expression becomes 
Jase = Yas = ees 6 eee (54) 
Consequently, as before, « and f are the roots of 
| 1 Yo — “Oy Ys = Yo == 0) i aber (55) 
| g Us — Yo Ys. — Y3 
Brg Ao Gi) ily a trig 
Similarly, if we have three indices, viz., p, q and r, we 
shall have 
Ym+1 = a2 + Lom + Mpm “Dey ee oe (56) 
which by subtraction reduces to 
ng = gma = Do™ + P+ Ree ee (57) 
so that a, 6 and y are the roots of 
1. to => Hh 9s - Ya Gaye =) 0) eae (58) 
S We = Ys ais Gomera 
E Yat Be Ye = YaeGer aaa 
CS 9s = ty. Ye = is Mee 
