759] 
ILLUSTRATION OF A THEOREM IN THE THEORY OF EQUATIONS. 
269 
If from these equations we eliminate /3, we obtain two equations in a, which it 
might be supposed would determine a uniquely; but, by what precedes, a is any 
root at pleasure of the cubic equation and can thus be determined only by the 
cubic equation itself, and it follows that any equation obtained by the elimination 
of ¡3 must contain as a factor the cubic function a 3 -7a + 6, and be thus of the form 
M (a 3 — 7a + 0) = 0, where M is a function of a; one result of the elimination is 
a 8 — 7a + 6 = 0, and every other result is of the form just referred to, i¥(a 3 —7a + 6) = 0 ; 
hence we have definitely a 3 — 7a + 6 = 0, viz. the roots of the equation M = 0 do not 
apply to the question. 
6 
In verification, observe that the first and second equations give a 2 — 7 = -, that 
is, a 2 — 6a + 7 = 0. To eliminate /3 from the first and third equations we first find 
a/3 2 + (4a 2 - 7) ¡3 + a 3 + 1 = 0, 
or say 
/ 7\ l 
/3 2 + i 4a j /3 + a 2 + - = 0, 
and combining herewith the first equation 
/3 2 + a/3 + a 2 — 7 = 0, 
we obtain 
that is, 
/3(3a-- +7+- = 0, 
ß = 
7a+ 1 
— 3a 2 + 7 ’ 
substituting in the first equation, 
(7a+ 1) 3 
+ a(7a + 1) (— 3a 2 + 7) 
that is, 
or, dividing by 3, 
which, in fact, is 
+ (a 2 — 7) (— 3a 2 + 7) 2 = 0, 
49 14 1 
-21-3+49 +7 
9 0-105 +343 -343 
9 0 - 126 - 3 + 441 + 21 - 342, 
3a 6 — 42a 4 — a 3 + 147 a 2 + 7a — 114 = 0, 
(a 3 - 7a + 6) (3a 3 - 21a - 19) = 0, 
of the form in question M(a 3 — 7a + 6) = 0. lhus a has any one at pleasure of the 
7a+ 1 
three values 1, 2, — 3, but a being known we have ¡3 — 
— 7a — 1 
7 = — a + 
— 3a 2 + 7 
3a 3 — 14a — 1 
-, and thence 
- 3a 2 + 7 ’ - 3a 2 + 7 
in particular, as a = l, then /3 = 2 and 7 = —3.