58 ON THE ALGEBRAICAL SOLUTION OF 



Hence, finally, substituting these values of a, 6, f in (1), we 

 find the identity 



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

 - (39750< 3 + 2763C 2 + 6270*+ 460) 3 + (19080* 3 + 14280* 2 

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

 Thus t=Q gives 



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

 while i 1 gives 



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

 By a similar process we obtain 

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



- (540 3 -456* 2 + 126t- 13) 3 + (540 3 -456 2 + 126<-9) 3 



-(864* 3 -816< 2 + 240^-21 ) 3 , 



and41 = -(246* 2 -264f+76) 3 +(1968i 3 -3096* 2 +1680<-310) 3 

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

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

 t=0, and t=l give respectively 



17==_43_22 3 +13 3 -9 3 +21 3 , 



= -64 3 + 266 3 - 197 3 + 20P-267 3 , 

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



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



