Taking for greater simplicity the cubic in the reduced form x^3 + qx -
r = 0, and assuming x = a + b, this will be a solution if only 3ab = q
and a^3 + b^3 = r, equations which give (a^3 - b^3)^2 = r^2 -
(4/27)q^3, a quadric equation solvable by radicals, and giving a^3 -
b^3 = [root](r^2 - (4/27)q^3), a 2-valued function of the
coefficients: combining this with a^3 + b^3 = r, we have a^3 = 1/2{r +
[root](r^2 - (4/27)q^3)}, a 2-valued function: we then have a by means
of a cube root, viz.
a = [root 3][1/2{r + [root](r^2 - (4/27)q^3)}],
a 6-valued function of the coefficients; but then, writing q = b/3a,
we have, as may be shown, a + b a 3-valued function of the
coefficients; and x = a + b is the required solution by radicals. It
would have been wrong to complete the solution by writing
b = [root 3][1/2{r - [root](r^2 - (4/27)q^3)}],
for then a + b would have been given as a 9-valued function having
only 3 of its values roots, and the other 6 values being irrelevant.
Observe that in this last process we make no use of the equation 3ab
= q, in its original form, but use only the derived equation 27a^3 b^3
= q^3, implied in, but not implying, the original form.
An interesting variation of the solution is to write x = ab(a + b),
giving a^3 b^3(a^3 + b^3) = r and 3a^3 b^3 = q, or say a^3 + b^3 =
3r/q, a^3 b^3 = (1/3)q; and consequently
3/2 4 3/2 4
a^3 = --- {r + [root](r^2 - --q^3)}, b^3 = --- {r - [root](r^2 - --q^3)},
q 27 q 27
i.e. here a^3, b^3 are each of them a 2-valued function, but as the
only effect of altering the sign of the quadric radical is to
interchange a^3, b^3, they may be regarded as each of them 1-valued; a
and b are each of them 3-valued (for observe that here only a^3 b^3,
not ab, is given); and ab(a + b) thus is in appearance a 9-valued
function; but it can easily be shown that it is (as it ought to be)
only 3-valued.
In the case of a numerical cubic, even when the coefficients are real,
substituting their values in the expression
x = [root 3][1/2{r + [root](r^2 - (4/27)q^3)}] + (1/3)q /
[root 3][1/2{r + [root](r^2 - (4/27)q^3)}],
this may depend on an expression of the form [root 3]([gamma] +
[delta]i) where [gamma] and [delta] are real numbers (it will do so if
r^2 - (4/27)q^3 is a negative number), and then we _cannot_ by the
extraction of any root or roots of real positive numbers reduce [root
3]([gamma] + [delta]i) to the form c + di, c and d real numbers; hence
here the algebraical solution does not give the numerical solution,
and we have here the so-called "irreducible case" of a cubic equation.
By what precedes there is nothing in this that might not have been
expected; the algebraical solution makes the solution depend on the
extraction of the cube root of a number, and there was no reason for
expecting this to be a real number. It is well known that the case in
question is that wherein the three roots of the numerical cubic
equation are all real; if the roots are two imaginary, one real, then
contrariwise the quantity under the cube root is real; and the
algebraical solution gives the numerical one.
The irreducible case is solvable by a trigonometrical formula, but
this is not a solution by radicals: it consists in effect in reducing
the given numerical cubic (not to a cubic of the form z^3 = a, solvable
by the extraction of a cube root, but) to a cubic of the form 4x^3 - 3x
= a, corresponding to the equation 4 cos^3 [theta] - 3 cos[theta] = cos
3[theta] which serves to determine cos[theta] when cos 3[theta] is
known. The theory is applicable to an algebraical cubic equation; say
that such an equation, if it can be reduced to the form 4x^3 - 3x = a,
is solvable by "trisection"--then the general cubic equation is
solvable by trisection.
18. A quartic equation is solvable by radicals, and it is to be remarked
that the existence of such a solution depends on the existence of
3-valued functions such as ab + cd of the four roots (a, b, c, d): by
what precedes ab + cd is the root of a cubic equation, which equation is
solvable by radicals: hence ab + cd can be found by radicals; and since
abcd is a given function, ab and cd can then be found by radicals. But
by what precedes, if ab be known then any similar function, say a + b,
is obtainable rationally; and then from the values of a + b and ab we
may by radicals obtain the value of a or b, that is, an expression for
the root of the given quartic equation: the expression ultimately
obtained is 4-valued, corresponding to the different values of the
several radicals which enter therein, and we have thus the expression by
radicals of each of the four roots of the quartic equation. But when the
quartic is numerical the same thing happens as in the cubic, and the
algebraical solution does not in every case give the numerical one.
It will be understood from the foregoing explanation as to the quartic
how in the next following case, that of the quintic, the question of
the solvability by radicals depends on the existence or non-existence
of k-valued functions of the five roots (a, b, c, d, e); the
fundamental theorem is the one already stated, a rational function of
five letters, if it has less than 5, cannot have more than 2 values,
that is, there are no 3-valued or 4-valued functions of 5 letters: and
by reasoning depending in part upon this theorem, N.H. Abel (1824)
showed that a general quintic equation is not solvable by radicals;
and _a fortiori_ the general equation of any order higher than 5 is
not solvable by radicals.
19. The general theory of the solvability of an equation by radicals
depends fundamentally on A.T. Vandermonde's remark (1770) that,
supposing an equation is solvable by radicals, and that we have
therefore an algebraical expression of x in terms of the coefficients,
then substituting for the coefficients their values in terms of the
roots, the resulting expression must reduce itself to any one at
pleasure of the roots a, b, c ...; thus in the case of the quadric
equation, in the expression x = 1/2{p + [root](p^2 - 4q)},
substituting for p and q their values, and observing that (a + b)^2 -
4ab = (a - b)^2, this becomes x = 1/2{a + b + [root](a - b)^2}, the
value being a or b according as the radical is taken to be +(a - b) or
-(a - b).
So in the cubic equation x^3 - px^2 + qx - r = 0, if the roots are a,
b, c, and if [omega] is used to denote an imaginary cube root of
unity, [omega]^2 + [omega] + 1 = 0, then writing for shortness p = a +
b + c, L = a + [omega]b + [omega]^2 c, M = a + [omega]^2 b + [omega]c,
it is at once seen that LM, L^3 + M^3, and therefore also (L^3 -
M^3)^2 are symmetrical functions of the roots, and consequently
rational functions of the coefficients: hence