INDETERMINATE CUBIC EQUATIONS 53 



v) 3 +(xi 

 x 2m r-p) 3 



are sufficient to give algebraical solutions of equation (A) for 

 the two cases n=2m and n=2m+l. In the former case the 

 roots will be functions of the 2m quantities x lt x 2 , . . ., x 2m , 

 and of the ra quantities X, p, . . ., p ; in the latter the roots 

 will be functions of the 2m+l quantities x lt x 2 , . . ., x 2m+l , 

 and of the ra+1 quantities X, jz, . . ., p. 



QUESTION 2. Solve the equation 



xy(x-y)=-kz 3 (1) 



knowing a particular solution, say xa, y=b, z=c so that 



a&(a-6)=Xc 3 (2) 



Put x=Xjr+a, yx z r+b, z=x 3 r+c, (3) 



and (1) on expansion and rearrangement becomes 



(x 1 x 2 



+ (ab x 1 x z +ab ax z +bx 1 3Xc 2 ic 3 )r=0 (4) 



Hence, making the coefficient of r vanish by taking 



x 3 (ab x^Xz+ab ax 2 +bxj)/3^c 2 , (5) 



equation (4) is satisfied by taking 



AX 3 XiX 2 (Xi X 2 ) 



- 9a6(o. - ft)(a 8 



_ 

 - 6) V - 3ab\5a 3 - 6a 2 6 + ^)x^ + 3a 2 6(2a 3 - Gab* + Sft^z,' - a'(26 - a) V 



on substituting the value of x 3 given by (5) and replacing Xc 3 

 by ab (ab) from (2). 



Hence, if we call the denominator of r, A, we find, finally, 



^y=(x 2 r+b)=b(2a-b) 3 (bx 1 -ax 2 ) 3 

 ^z=(x 3 r+c)=-(a+b)(2a-b)(2b-a)(bx 1 -ax 2 ) 3 



which, it is to be remarked, is not an algebraical solution, but 



