MR. MURPHY ON THE ANALYSIS OF THE ROOTS OF EQUATIONS. 165 
To this end let 2 x 1 — x 2 — x 3 — A, and x 2 — ^3 = 6, then 
2 oc 2 - - x 3 — \ ( 3 B - A) 
2 x 2 x 2 — . ( 3 13 — A) 
therefore 
(2 x 2~~ (2 ^2 ' ^1 #3) (2 *% — — ^2) — "g" (A” — 9 A B 2 ). 
Again 
x \ ~ x 2 = 4 ( A ~ B ) 
x l - x 3 = 4" ( A + B ) 
therefore 
■ («1 - **) (*1 - *3) fe - %) = “~^~(A 2 B - B 3 ) 
the total quantity under the first cubic surd thus becomes 
Y { A 3 + 3 A 2 . (B ✓ - 3) + 3 A . (B ✓ - 3) 2 + (B V - 3) 3 }, 
the cube root of which is-^{A + B y/ — 3}, and the actual root of the second surd 
is similarly -5- {A — B s/ — 3}. But 
i(A + B ✓ — 3) = 
= x 1 -f- x 2 0 2 + x 3 6. 
And 
Y (A — B */ — 3) = x 1 + x 2 6 + x 3 6 ; 
the formulae of this solution become then those of Vaudermonde, viz. 
3 x Y = -J- <r 3 ) -J- (x l x 2 8 1 x 3 6) -f- (j?j + x 2 & + x 3 Q 2 ) 
3 # 2 = (<^2 -j- a? 2 + #3) + (^1 $ + x 2 + <r 3 0 2 ) + (x { d 2 + x 2 + x 3 0) 
3 x 3 — (#2 + x 2 + x 3 ) + (#1 + x 2 0 -f x 3 ) + (x l 6 + x 2 Q 2 + x 3 ), 
which contain a complete verification. 
17- The two constituent parts a' a" of the roots of a cubic equation have been re- 
solved into factors by observing the relations established between the roots by their 
evanescence ; another mode exists for forming the same quantities by elimination 
between the proposed equation and its first and second derived equations. 
Let the given equation x 3 + a x 2 + b x + c — 0 be represented by <p ( x ) = 0, 
and the first derived, viz. 3 x 2 + 2 a x + b = 0 by <p' (x) =0, the second derived 
6 x -f 2 a = 0 by <p" (x) = 0. 
