2. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,
x³ + Ax² + Bx + C = 0,
where A, B and C denote known quantities, either positive or negative.
This equation may be transformed into another in which the second term is wanting by the substitution x = y − A/3. This transformation is a particular case of a general theorem. Let xn + Axn−1 + Bxn−2 ... = 0. Substitute x = y + h; then (y + h)n + A (y + h)n−1 ... = 0. Expand each term by the binomial theorem, and let us fix our attention on the coefficient of yn−1. By this process we obtain 0 = yn + yn−1(A + nh) + terms involving lower powers of y.
Now h can have any value, and if we choose it so that A + nh = 0, then the second term of our derived equation vanishes.
Resuming, therefore, the equation y³ + qy + r = 0, let us suppose y = v + z; we then have y³ = v³ + z³ + 3vz (v + z) = v³ + z³ + 3vzy, and the original equation becomes v³ + z³ + (3vz + q) y + r = 0. Now v and z are any two quantities subject to the relation y = v + z, and if we suppose 3vz + q = 0, they are completely determined. This leads to v³ + z³ + r = 0 and 3vz + q = 0. Therefore v³ and z³ are the roots of the quadratic t² + rt − q²/27 = 0. Therefore
| v³ = | −½ r + √(1⁄27 q³ + ¼ r²); z³ = −½ r − √(1⁄27 q³ + ¼r²); |
| v = | 3√{−½ r + √(1⁄27 q³ + ¼ r²) }; z = 3√{ (−½ r − √(1⁄27 q³ + ¼ r²) }; |
| and y = | v + z = 3√{−½ r + √(1⁄27q³ + ¼ r²) } + 3√{−½ r − √(1⁄27 q³ + ¼ r²) }. |
Thus we have obtained a value of the unknown quantity y, in terms of the known quantities q and r; therefore the equation is resolved.
3. But this is only one of three values which y may have. Let us, for the sake of brevity, put
A = −½ r + √(1⁄27 q³ + ¼ r²), B = −½ r − √(1⁄27 q³ + ¼ r²),