An introduction to the modern theory of equations, by Florian Cajori.

GALOIS lIESOLVENT BY ADJ UzNCTION

177

0(y) = (y- y - I )( - 2 (a - Ct)2, (y) -?y(a -+ el) + au + c 12
- 2 aa =- - 8, 0'(y) 2 y. Hence a = 4(11)/p'J(M) =- 4/lL. The
correctness of this result is easily shown.
Ex. 2. For the equation x2 + ax + b 0, having the group P= 1,
(aac), find a3 - /12 as a function of a in (1).
Take Q = 1, M-1 a, 31' = cc - ((, theln 4(y) = (3 ab + 2 b - a2 - a3)?
+ 3 ab + 2 b2 - 2b - a3, 0(sQ) - 2 y + a. Hence
3' = [(3 ab + 2 b - a2 - a3) — + 3 ab - 2 -a -] - a(2 1+ a).
Ex. 3. Find the value of [co, a]3 for the cubic x3 4- ax2 + at2 + a3 = 0
in terms of the alternating function ((a - al) (( - a2) (a -- (a2) = D.
Let 1 = V/D, then 1 M1= - VD.
We have M' = [w, a]3, 1'lD [C2, a]3, p(y) Y2 - D,
4,(y) -y(,I' + 3i') -t+ VD(31 - M3). By ~ 71, Ex. 15,
M' + M' = - 2 a13 + 9 (aa1 - 27 a3. We find 1 - Mtl = - 3 i VD,
$(3M) _ VD(- 2 ait + 9 (ala2 - 27 a3 - 3 i V3 D),
'(31) _2 V/D, 3i =- -(-2 a13 + 9 a1a2 - 27 a3 -3 i /v'3D).
See also the solution in ~ 173.
Ex. 4. For the quartic X4 + 4 blx3 + 6 b.2 +- 4 b3C + b4 - 0, find the
value of M31 _ (a + ao) (a(l + aC) in terms of 31, where
16 i_ (a - a + a2 - a3)2.
Both M and M1 belong to the group G8(4). Notice that 31 is a root of
the cubic III, ~ 62. See also ~ 169. Hence that cubic is 0(y) =0. We find
162 p(y) 162 (31t + 431'- + 31'l1)y2 - 16({Jil + 31i}Mt1 + {31 + I i}3It
4+ 3I + 41- 31'}11) y + il11/! 31' + 1-nM 313JIll31 + 31I4tM'11t
= 162 2 Sat1a2. y2 - 16 (4 S2ac1 ~ Za2 - 8 Za2sCaao)?
+ (2 a5al1 - 6 ja4a1a(. 4- 4 Sa3al2a2 - 4 a3a13 - 4 Sa2al2 a2a3).
In Ex. 16, ~ 71, the values of the symmetric functions occurring here are
given.
Ex. 5. Complete the computation in Ex. 4 for the special quartic
X4 - 6 2 +- 4+ 1 =. We obtain ic(y) _12y - 16y -3,
~(/)-23$Y22Y 1 ___401 M+ 11
Cy>) _ y + 3 + 2 - M --- 4 ~....., ( r 4 '3.11'2 + 6 31 - 2
N