INDETERMINATE CUBIC EQUATIONS 51 

 To make the coefficient of r vanish we may take 



*o= (^2*1+ V***+ f^s) A 2 (5) 



and equation (4) is then satisfied by taking 



r=3(xs 1 2 + [*z 2 2 -f*:r 3 2 -XV)/(V-*i 3 -*2 3 -*3 3 ) (6) 



Substituting the value of x from (5) in (6) we derive 



If we nowput A-(X 2 a; 1 +[ji 2 a; 2 +(x 2 a; 3 ) 3 -X 6 (a; 1 3 +a; 2 3 +a:3 3 ) (8) 

 equations (3) take the form 



A P = 



+3x(X 2 z 1 + ^x z + [* 2 * 3 )Sx V+ I* 

 A P= 



A P 2 =A(a; 2 r+ ! i)=(x(X 2 a; 1 + !J i 2 a; 2 +( A 2 X3) 3 -XV(a; 1 3 +^2 3 +a; 3 3 ) 



+ 3x 

 A P= 



(9) 



As equation (1) is homogeneous, (9) is the integralised 

 form of its algebraical solution and presents the roots as rational 

 functions of five variables x lf x z , x 3 , X, ;/. ; of the third degree 

 in x lt x z , x a , the ninth in x lt x z , x a , jx, and the tenth in x lt x z , 



As a numerical example, a? 1 =a; 2 =X = l, a; 3 =p(.=2 gives, on 

 removal of the common factor, 3 3 +4 3 +5 3 =6 3 , the lowest 

 solution which exists. 



(ii) Let -=4, so that we have to solve 



P 3 =P 1 3 +P 2 3 +P 3 3 +P 4 3 (!') 



Here we may take as our particular solution 



P =P 1= X, P 2 =-P 3 =[z, P 4 =0 (2 y ) 



Making then the substitutions 



P =x r+ X P =x r-\- X P x r-\- u. P =x T y. PA=X*T, (3') 

 equation (!') takes the form 



(x r+ X) 3 = (XjT+ X) 3 + (x z r+ (i) 3 + (x 3 r [x) 3 + (x^r) 3 



