CUBIC EQUATIONS. 575 



answer to the question, whether any given number N is a prime or not ? The 

 solution is complete, but the actual calculation, from its length, in most cases, 

 seems impossible. 



In the solution of any algebraic problem, a knowledge of arithmetic may be 

 presupposed. As soon as the required operation is indicated, we conclude that 

 arithmetic has performed it, by its often tedious, but steady and unerring rules ; 

 and the task of the algebraist has ended when he has shown how his problem 

 falls within the scope of those rules. 



I come now to the more immediate subject of this paper. 



(1.) Let the proposed cubic be 



x 3 + qx — r = 



which will become y 3 + qy — r = 



and z 3 + qz — r = 



if the other roots are written instead of x. Adding the three equations together. 



we have 



{x 3 + y 3 + z 3 ) + q(x + y + z) - 3r = 



But x + y + z — .-. x 3 + y 3 + z 3 = 3»- 



or in my notation Sx 3 = 3r 



(2.) Since x 3 — - qx + r 



and y 3 = — qy + r 



.-. their product x 3 y 3 = q 2 xy — qr(x + y) + r 2 



Whence x 3 y 3 + yh 3 + z 3 x z = $x 3 y 3 = q 2 Sxy — 2qrSx + 3r 2 



For, the process being the same with regard to each of the roots, the sum of the 

 equations can be inferred from any one of them. 



The above value of S# 3 ?/ 3 reduces itself (since Sxy = q, S% — 0) to 



Sx 3 y 3 = q 3 + 3r 2 



(3.) There are six products of the form x 2 y. It is evident that the sum of all 



six is equal to 



xy(x + y) + xz(x + z) + yz(y + z) 



But x + y — — z, :. the first term becomes — xyz or — r. Similarly for the 

 two other terms ; .-. the sum of six products like x 2 y — — 3r. But here a more 

 important question arises. If we separate the six products into two groups of 

 three each, taken in order, namely, 



m = x 2 y + y 2 z + z 2 x 

 n = y 2 x + z 2 y + x 2 z 



The first may be conveniently represented by &>x 2 y = m, and the second by 

 §y-x = n, and we are led to the inquiry, What are the values of m and n, con- 

 sidered separately ? This problem may be solved as follows : — 



