io6 A NEW SOLUTION 



denominator of the radical, I find that it is always divisible by 

 M, the quantity in the numerator : and that thus the formulas 

 may be exhibited in another fhape, having this advantage, that 

 it will introduce fmaller numbers in the arithmetical operations 

 requiute for computing a and b. 



I write QxM=3(BN— 3 CM)M-2A(BN — 3CMX3N— AM) 

 -f-B(3N — AM) 2 , and evolving by actual multiplication 

 QJK M = 3B 2 . N* — 1 SBC. MN -f 27C 2 . M 2 

 -f 6AB. N 2 — 2A-B. MN 



— 1 SAC. MN + 6A 2 C. M 2 

 -f 9B. N 2 — 6AB.MN + BA 2 . M 2 

 Now, the coefficient of N 2 is, 3B 2 -f 6AB -f 9B = 

 3B (B -f 2 A + 3) — 3B X M : dividing therefore by M, we get, 

 Qj=3B XN'» (18BC + 2 A 2 B + 18AC + 6AB) XN 

 + (27C 2 -f 6A 2 C + BA 2 ) X M. 



And if for N and M we fubftitute their values (A -f 2B -f- 3C) 

 and (3 + 2A + B), we fhall find, 



Q_=: 12B 5 +81C 2 — 54ABC+i2A 3 C — 3A 2 B* 

 all the other terms deftroying one another, except thefe five. 



All the terms in this value of Q^ being divifible by 3, 1 

 change Q^and put 



Q= 4 B'+27C i — i8ABC + 4 A 3 C--A 2 B 2 

 or, Qj= (4B 3 + 27C 2 ) — 2 AC ( 9 B — A 2 ) -f A 2 (2AC — B 2 ) 

 And fince 3QM. is now equal to the denominator of the radical 

 in the preceding formula; for a and £, we have the following 

 new formulas, 



4. BN-3CM . 

 a ~~ •T3Q. 



, , + aN-AM 



What 



