70 ON THE ALGEBRAICAL SOLUTION OF 



On expansion and rearrangement, (2) becomes 

 4(Xaz 1 3 + [ xfo; 2 3 -Xca; 1 3 - ( idz 2 3 )r 3 +6(Xa 2 z 1 2 + n&^-Xc 2 ^ 2 



-\td?xf)r a +tyla*x 1 +v.lfar-l(Px l --\iLcPxjr=0 (4) 



To make the coefficient of r vanish we take 



Xl =u.(d 3 -b 3 )x z ma 3 -c 3 ). (5) 



Equation (4) is then identically satisfied by taking 



3[X(c'-o a )g 1 + j*(d 2 -& 2 ):r 2 2 J 



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

 Hence 



x r , - 



2[t A 2 (a-c)p 3 -6 3 ) 3 +X 2 (6-rf)(a 3 -c 3 ) 3 ] 

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



Also 



r r+?? _3x(a 3 -c 3 )[|x(c 2 -a 2 )(rf 3 -6 3 ) 2 +X(^-6 2 )(a 3 -c 3 ) 2 ] , 



=(d-&)(3d+6)(a 3 -c 3 ) 3 X 2 +3(c 2 -a 2 )(a 3 -c 3 ) 



(rf 3 -6 3 ) 2 X(ji+26(a-c)(d 3 -6 3 ) 3 (A 2 . 

 Also 



* r , c _3ix(rf 3 -6 3 )[^(c 2 -a 2 )(rf 3 -6 3 ) 2 +X(^-& 2 )(a 3 -c 3 ) 2 



Also 



,_-ix---- rf 



