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.