# 1911 Encyclopædia Britannica/Equation/Cubic Equations

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1. Cubic equations, like all equations above the first degree, are divided into two classes: they are said to be pure when they contain only one power of the unknown quantity; and adfected when they contain two or more powers of that quantity.
Pure cubic equations are therefore of the form x$^3$ = r; and hence it appears that a value of the simple power of the unknown quantity may always be found without difficulty, by extracting the cube root of each side of the equation. Let us consider the equation x$^3$ − c$^3$ = 0 more fully. This is decomposable into the factors x − c = 0 and x$^2$ + cx + c$^2$ = 0. The roots of this quadratic equation are 1/2 (−1 ± √−3) c, and we see that the equation x$^3$ = c$^3$ has three roots, namely, one real root c, and two imaginary roots 1/2 (−1 ± √−3) c. By making c equal to unity, we observe that 1/2 (−1 ± √−3) are the imaginary cube roots of unity, which are generally denoted by ω and ω$^2$, for it is easy to show that (1/2 (−1 − √−3))$^2$ = 1/2 (−1 + √−3).
2. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,
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 x$^n$ + Ax$^{n−1}$ + Bx$^{n−2}$ ... = 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 y$^{n−1}$. By this process we obtain 0 = y$^n$ + y$^{n−1}$(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 y$^3$ + qy + r = 0, let us suppose y = v + z; we then have y$^3$ = v$^3$ + z$^3$ + 3vz (v + z) = v$^3$ + z$^3$ + 3vzy, and the original equation becomes v$^3$ + z$^3$ + (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 v$^3$ + z$^3$ + r = 0 and 3vz + q = 0. Therefore v$^3$ and z$^3$ are the roots of the quadratic t$^2$ + rt − q$^2$/27 = 0. Therefore
| v$^3$ = | −1/2 r + √(1/27 q$^3$ + 1/4 r$^2$); z$^3$ = −1/2 r − √(1/27 q$^3$ + 1/4r$^2$); |
| v = | ∛{−1/2 r + √(1/27 q$^3$ + 1/4 r$^2$) }; z = ∛{ (−1/2 r − √(1/27 q$^3$ + 1/4 r$^2$) }; |
| and y = | v + z = ∛{−1/2 r + √(1/27q$^3$ + 1/4 r$^2$) } + ∛{−1/2 r − √(1/27 q$^3$ + 1/4 r$^2$) }. |
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
| and put | α = 1/2 (−1 + √−3), |
| | β = 1/2 (−1 − √−3). |
Then, from what has been shown (§ 1), it is evident that v and z have each these three values,
To determine the corresponding values of v and z, we must consider that vz = −1/3 q = ∛(AB). Now if we observe that αβ = 1, it will immediately appear that v + z has these three values,
v + z = α∛A + β∛B,
which are therefore the three values of y.
​The first of these formulae is commonly known by the name of Cardan’s rule (see Algebra: History).
The formulae given above for the roots of a cubic equation may be put under a different form, better adapted to the purposes of arithmetical calculation, as follows:—Because vz = −1/3 q, therefore z = −1/3q × 1/v = −1/3 q / ∛A; hence v + z = ∛A − 1/3 q / ∛A: thus it appears that the three values of y may also be expressed thus:
y = α∛A − 1/3 qβ / ∛A
See below, Theory of Equations, §§ 16 et seq.