70 ON THE ALGEBRAICAL SOLUTION OF 

 On expansion and rearrangement, (2) becomes 



)r=0 (4) 



To make the coefficient of r vanish we take 



a ; 1 = (x (d3-&3)a; 2 /X(a 3 -c 3 ). (5) 



Equation (4) is then identically satisfied by taking 



3|>(c 2 -aW 



on substituting for x l its value given by (5). 

 Hence 



~ r , ff _ 



1 2[(x 2 (a-c)((f 3 -6 3 ) 3 +X 2 (&-cl)(a 3 -c 3 ) 3 ] 



i.e. if we write A for 2|> 2 (a-c)(d 3 -6 3 ) 3 + X 2 (6-rf)(a 3 -c 3 ) 3 ] 



Also 



r r ,7 f _3M 3 -c 3 )[tx(c 2 -a 2 )(^-6 3 ) 2 +X(^-6 2 )(a 3 -c 3 ) 2 ] , 



Also 



rr r , r _ 



23332333 



[X 2 (d 3 -6 3 ) 3 [3c 2 -3a 2 -f 2ac-2c 2 ] 



Also 



, ----- , 



