372 Proceedings of Royal Society of Edinburgh. [sess. 
1 , Cl. 2 , j £iq) = (ctj + ^2 + ^3 + ^4X^1 ~ Ct^-^r Ct^ — (X 4 ) 
1 ...... s dq) = (cq + + $3 + $4 + oq) 
^2 ^3 
— oq 
a b - a-j 
$3 d/j 
a A 
a. 
4 2 
C<3 
O /3 6 X 4 £X 4 (3q $q £]q 
a, 
a K 
a. 
aq aq — a 2 
$q ^q ^q ^2 
a 2 ~ a 3 
Ct-^ (t 2 
$2 $q Cq U 4 
(6) In the case where w = 6 two factors are as yet unknown, 
viz., the two corresponding to the factors x 2 + x-\-\, 0C> + 1 of 
x 6 — 1. For finding the first of the two, the equations concerned 
are 
a x x b + a 2 x^ + a 3 x s + a^x 2 + a b x + a 6 = 0 
x 2 + x + 1 = 0 
With the help of the second equation the term ayp can he 
removed from the first, the result being 
(ci 2 - aq)# 4 + (a 3 — a 1 )x 3 + a^x 2 + a b x + a Q = 0 . 
From this equation, again, the term (a. 2 - aq)^ 4 can he removed in 
the same way with the result 
(a 3 - a 2 )x 3 + (a i -a 2 + af)x 2 + ci b x + a 6 — 0 , 
and this process can clearly he continued until there occurs in the 
resulting equation no power of x higher than the first. Doing so, 
we have finally 
(a b - a 4 + a 2 - a^)x + (a 6 - a 4 + a 3 -a 1 ) = 0 , 
and, by cyclical substitution, 
(a 6 -a 5 + a 3 - a 2 )x + (cq - a b + a A - a 2 ) = 0 , 
from which two equations there results the eliminant 
a b — cx 4 + a 2 — a 4 cq — <x 4 + a 3 — oq 
ci 6 - a b + a 3 - a 2 a 1 - a 5 + a 4 - a 2 , 
or, in persymmetric form, 
cl 3 — & 4 + (Xg — cq cq — a b + cq — ci 2 
a 4 — cc b + cq — a 2 a b — cq + ci 2 — ct 3 , 
and this is the desired factor. 
By combining the various subtractions in this process, the work 
