835] 
299 
835. 
ON CARDAN’S SOLUTION OF A CUBIC EQUATION. 
[From the Messenger of Mathematics, vol. xiv. (1885), pp. 96, 97.] 
It is interesting to see how the solution comes out when one root of the equation 
is known. Say the equation is a? + qx — r = 0, where a s — qa — r = 0, that is, r = a 3 4- qa. 
Solving in the usual manner, we have 
whence 
and thence 
(y 3 
or say 
and therefore 
x — y + z, y 3 + z 3 — r + {y + z) (f>yz + q) = 0, 
y 3 + z 3 = r, 
yz = -iq ; 
— z 3 ) 2 = r 2 + f Y q 3 , = a 6 + 2ga 4 + fa- — f 7 q 3 , = (a 2 + f q) {a 2 + fg) 2 ; 
y 3 z 3 = (a 2 + lq) f(a 2 + §q) ; 
8y 3 = 4a 3 4- 4qa + (4a 2 + |g) f(a 2 + §q), = [a + \J{a 2 + |g)} 3 , 
8^ s = 4a 3 + 4qa — (4a 2 + §q) f{a 2 + |q), = (a — f{a 2 + fg)} 3 ; 
where observe that the essential step is the expression of the irrational functions as 
perfect cubes: that the functions are the cubes of a ± f(a 2 + fg) respectively is seen 
to be true; but if we were to attempt to find a cube root a + /3 f (a 2 + fg) by an 
algebraical process, we should be thrown back upon the original cubic equation. 
Writing then w for an imaginary cube root of unity, we have 
. 2y = (1, to or ft> 2 ) [a + f(a 2 + fg)}, 
2z = (1, co 2 or <w ) {a — f{a 2 + -fg)} ; 
and then 
x = y + z — a, or — — ^a ± l — &r) f(a 2 + fg), 
where w — a> 2 — i f 3; the last two roots are of course the roots of the quadric equation 
x 2 + ax + a 2 + g = 0, which is obtained by throwing out the factor x — a from the given 
equation x? + qx — r = 0. 
Cambridge, Sep. 17, 1884. 
38—2