342 W. Spottiswoode. On RusseUs Integrals. [Feb. 9, 



we have, omitting as before the last equation, viz,, that connecting g 

 and 8 — 



€1— u\ 3 , 

 b = oiXPfi, 



hd = 3aXfJLV + 2ayii 3 + 3(3\jU, 



5e= «\i> 2 + 4*iyi 2 + 2/3\v + 4/3^ 3 + 7\, 



Combining these, it will be found not only that, as in the case of the 

 quartic, 



Scju, — dX=2ac\fiP ; 

 but also that 4/t 2 . hd— 3fiX . 5e + 3\ 3 /=8a,a 5 , 



from which both /3 and v have simultaneously disappeared. Hence 

 we have for the elimination of «, X, /t, the four equations : — 



a— ot\?, 



& = #\ 2 £t, 



3c/i—d\=2<z\iu?, 

 20c^ 2 - 1 5e V + 3/X 2 = 8^ 5 , 

 from which it is easy to derive the two final conditions — 



a?d-3abc + 2b*=0, 

 and 20a-bM-lba%e + 3a 4 f-8b^=0. 



It would perhaps be difficult to obtain the results for the general 

 case by the present method ; but if the question be regarded from a 

 different point of view, we can see not only how many conditions will 

 exist in general, but also what will be their form ; and, moreover, we 

 can actually obtain them in any particular case. 



The problem is in fact identical with that of finding the conditions 

 under which an equation of the degree 2n can be solved by means of 

 an equation of the degree n. To answer this, we have merely to sub- 

 stitute in the equation 



(a, b, . . .)(», iy*=o 



x + w for x, and equate to zero the coefficients of the odd powers of x. 

 This will give n equations, from which we can of course eliminate w 

 in lways; and the results of the eliminations will consequently 

 be the conditions sought, n — 1 in number. Applying this to the case 

 of the sextic, we have 



O, b, . . .)O + ^) 6 =0; 



