72 ON THE ALGEBRAICAL SOLUTION OF 



Again, putting d=Q, we have that if Xa 4 + fx& 4 =Xc 4 , then also 



=X[2c(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +36 4 (a 2 +ac+c 2 )(a 3 -c 3 )6 4 X{x 



+ [x[36(a 2 +ac+c 2 )(a 3 -c 3 ) 2 X 2 +3(c+a)(a 3 -c 3 )6 5 X(ji] 4 . 



Thus to solve P 1 4 +5P 2 4 =P / 1 4 + 5P V since ! 4 +5-2 4 =3 4 +5-0 4 , 

 we derive 



2. The equation XP X 4 + [AP 2 4 = XP\ 4 + [zP' 2 4 + ^ 

 is always soluble whatever be the value of I, provided a 

 particular solution be known. For by putting P^ 

 P 2 =uv, P\=xy, and P' z =u+v, it becomes 



which is an indeterminate cubic in x and u. We proceed to 

 solve the case where ^^(P'g 4 P 3 4 ). 



QUESTION 2. Solve the equation 



xpjM- ^P 2 4 + vP^xPY+^py+vP's 4 * (i) 



knowing a particular solution, say 



Xa 4 + (Ji6 4 + z/c 4 = Xd 4 + [xe 4 + p/ 4 (2) 



Here we may assume for equation (1) 



(3) 



which on expansion and rearrangement becomes 



=0 (4) 



To make the coefficient of r vanish we must take 



