Equation of the Fifth Degree, 34f5 



5S*2 = J^i + a^^2 + '^^'^a + *'*'^4 + "^^ 

 Sd\ = a?i + «*^2 + «*^3 + <''^4 + ^^^, 

 56*4 = iTi + a^3 + ^^^3 + *^^4 + "''^S' 



Let 5i, figj 635 ^4 be the roots of 



fl4 + Mg3^Nfl2 + Pfl + Q=0. 

 Then 



-M=0, +624-63+64; 



to find which I employ 



(m + u + to + 3 + ^)5 = 2(«^) + 5X Ku) + 1 02(w3u^) + 202 (M^tw) 

 + 302(w^u^w) + 602(m^«W2) + l^Owmzt, 



collecting the terms separately, and reducing by means of 



l+ot + a%+a3 + a^ = 0, 0:1 + 5-2 + ^3 + ^4 + ^-5=0, 



and also 



X\ - X^X^ = - x\{Xc^ + 0^3 + ^4 + ^bli 



a^lx^x^ + x^x^) = - x\xlx^ - x\xlx^ - x\x\x^ - x\xlx^ 

 —x\[XciX^X^ + Xc^XgXr^ + Xc^X^X^ + x^x^x^, &c. 



We thus find 



5^(3i + fl2 + S3 + M = 192«)-102(o;^)X«)-202(^^^2^3) 

 — 1 302(^^4^3) ~ 4-OS (^?a;2a:'3^4) + 4i^0x^Xc^QX^x^ + 250 



+ 5;^o-«or, + aX^i + ^K^4 + ^^X) • 



The first six terms of the second member are all given, 

 being symmetrical functions of a7j, a?2j &c. Let their sum be 

 R; then, putting — M for 5)(6i), the above will be 



-5^M-R=s260(a?2ar2a74 + x2a?2a?j + ). 



Or if we make 



5^M + R _ 



250 ~'^* 

 it will become 



-<^=x\xlx^^x\xlx^-\- ' 



(6.) 



Now if we make a?i, x^ change places in the second member of 

 the last, then a'^, ajg, and o-g* ^3> &c., we shall find that it has 

 six different values, as stated by Lagrange. Thus 



