21. It is to be remarked, in regard to the question of solvability by radicals, that not only the coefficients are taken to be arbitrary, but it is assumed that they are represented each by a single letter, or say rather that they are not so expressed in terms of other arbitrary quantities as to make a solution possible. If the coefficients are not all arbitrary, for instance, if some of them are zero, a sextic equation might be of the form x^6 + bx^4 + cx^2 + d = 0, and so be solvable as a cubic; or if the coefficients of the sextic are given functions of the six arbitrary quantities a, b, c, d, e, f, such that the sextic is really of the form (x^2 + ax + b)(x^4 + cx^3 + dx^2 + ex + f) = 0, then it breaks up into the equations x^2 + ax + b = 0, x^4 + cx^3 + dx^2 + ex + f = 0, and is consequently solvable by radicals; so also if the form is (x -a)(x - b)(x - c)(x - d)(x - e)(x - f) = 0, then the equation is solvable by radicals,--in this extreme case rationally. Such cases of solvability are self-evident; but they are enough to show that the general theorem of the non-solvability by radicals of an equation of the fifth or any higher order does not in any wise exclude for such orders the existence of particular equations solvable by radicals, and there are, in fact, extensive classes of equations which are thus solvable; the binomial equations x^n - 1 = 0 present an instance. 22. It has already been shown how the several roots of the equation x^n - 1 = 0 can be expressed in the form cos 2s[pi]/n + i sin 2s[pi]/n, but the question is now that of the algebraical solution (or solution by radicals) of this equation. There is always a root = 1; if [omega] be any other root, then obviously [omega], [omega]^2, ... [omega]^(n - 1) are all of them roots; x^n - 1 contains the factor x - 1, and it thus appears that [omega], [omega]^2, ... [omega]^(n - 1) are the n - 1 roots of the equation x^(n - 1) + x^(n - 2) + ... x + 1 = 0; we have, of course, [omega]^(n - 1) + [omega]^(n - 2) + ... + [omega] + 1 = 0.