58 ON THE ALGEBRAICAL SOLUTION OF 



Hence, finally, substituting these values of a, &, in (1), we 

 find the identity 



5= (1590* 2 + 660*+ 62) 3 + (39750* 3 + 27630* 2 + 6270*+458) 3 

 - (39750* 3 + 27630* 2 + 6270*+460) 3 + (19080* 3 + 14280* 2 

 + 3432*+ 264) 3 - (19080* 3 + 14280* 2 + 3432*+ 259) 3 . 

 Thus *=0 gives 



5=62 3 +458 3 -460 3 +264 3 -259 3 , 

 while t= 1 gives 



5 = 992 3 - 17932 3 + 17930 3 - 7968 3 + 7973 3 . 

 By a similar process we obtain 

 17=-(108* 2 -48*+4) 3 +(864Z 3 -816Z 2 +240-22) 3 

 -(540* 3 -456* 2 + 126*- 



1 = -(246* 2 -264*+76) 3 +(1968* 3 -3096* 2 +1680*-310) 3 

 -(1968* 3 -3096* 2 +1680*-311) 3 +(738* 3 -1284* 2 

 + 762*- 157) 3 - (738* 3 - 1284* 2 + 762*- 159) 3 . 

 *=0, and *=1 give respectively 



17 = -4 3 - 22 3 + 13 3 -9 3 + 21 3 , 



= -64 3 +266 3 -197 3 +201 3 -267 3 , 

 and 41 = -76 3 -310 3 +311 3 -157 3 +159 3 , 



= -58 3 +242 3 -241 3 +59 3 -57 3 . 



