Reverting to the before-mentioned particular equation x4 + x³ + x² + x + 1 = 0, it is very interesting to compare the process of solution with that for the solution of the general quartic the roots whereof are a, b, c, d.
Take ω, a root of the equation ω4 − 1 = 0 (whence ω is = 1, −1, i, or −i, at pleasure), and consider the expression
(a + ωb + ω²c + ω³d)4,
the developed value of this is
| = | a4 + b4 + c4 + d4 + 6 (a²c² + b²d²) + 12 (a²bd + b²ca + c²db + d²ac) |
| +ω | {4 (a³b + b³c + c³ + d³a) + 12 (a²cd + b²da + c²ab + d²bc) } |
| +ω² | {6 (a²b² + b²c² + c²d² + d²a²) + 4 (a³c + b³d + c³a + d³b) + 24abcd} |
| +ω³ | {4 (a³d + b³a + c³b + d³c) + 12 (a²bc + b²cd + c²da + d²ab) } |
that is, this is a 6-valued function of a, b, c, d, the root of a sextic (which is, in fact, solvable by radicals; but this is not here material).
If, however, a, b, c, d denote the roots r, r², r4, r³ of the special equation, then the expression becomes
| r4 | + r³ + r + r² + 6 (1 + 1) | + 12 (r² + r4 + r³ + r) |
| + ω {4 (1 + 1 + 1 + 1) | + 12 (r4 + r³ + r + r²) } | |
| + ω²{6 (r + r² + r4 + r³) | + 4 (r² + r4 + r³ + r) } | |
| + ω³{4 (r + r² + r4 + r³) | + 12 (r³ + r + r² + r4) } |
viz. this is