Finding the roots of a cubic equation

Question

I trying to find the roots of the equation

$$ax^3 + bx^2 + cx + d = 0$$

By using some changes of variable (which does not really matter now) I was able to rewrite this equation as

$$z^3 - \frac{\Delta_{_{0}}}{3a^2}z+\frac{\Delta_{_{1}}}{27a^3} = 0$$

I used then the Vieta's substitution

$z = w + \frac{\Delta_{_{0}}}{9a^2w}$

to obtain the following equation:

$$(w^3)^2 + \frac{\Delta_{_{1}}}{27a^3}w^3 + \frac{\Delta_{_{0}}^{3}}{729a^6} = 0$$

This is a quadratic equation on $w^3$ with roots:

$$ w^3 = \frac{1}{27a^3}\,\frac{-\Delta_{_{1}} \pm \sqrt{\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3}}}{2}$$

There are then six possible values for $w$. Denoting

$$ C=\sqrt[3]{\frac{\Delta_{_{1}}+\sqrt{\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3}}}{2}} \text{ and } \bar{C}=\sqrt[3]{\frac{\Delta_{_{1}}-\sqrt{\Delta_{_{1}}^{2}-4\Delta_{_{0}}^{3}}}{2}}$$

they are

$$ w = -\frac{\mu_{_{k}}}{3a}C \text{ and } w = -\frac{\mu_{_{k}}}{3a}\bar{C} $$

where $\mu_{_{k}} = \left\{ 1, \frac{-1 \pm i\sqrt{3}}{2} \right\}$ for $k=1,2,3.$

Finally... my point is: I can substitute any of the six $w$'s into the Vieta's substitution formula and obtain six values for $z$ when only three should exist. What am I missing?

UPDATE

Since $u_{_{k}}^{3} = 1$ there are three equivalent solutions for $C$ and three for $\bar{C}$ and so there are only two (possibly) distinct solutions for $w$:

$ w = -\frac{1}{3a}C \,\,\,\,\,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\,\,\,\,\, w = \frac{1}{3a}\bar{C}$

which will give me only two possible values for z. A short while ago I had three solutions too many and now I'm one short. The confusion goes on...

Thank you very much.

Answer 1

Regardless of whether all six are distinct (and I suspect not), it should be clear that Vieta's substitution does not typically have a unique $w$ that yields a given $z$. In that substitution, $w$ is a quadratic function of $z$.

Answer 2

The $w$ are conjugate in pairs, and the values of $z$ remain the same for a pair of conjugates.

Answer 3

First find the solutions of ${x^3} + px + q = 0$.

Comparing ${x^3} + px + q = 0$ with ${x^3} + {a^3} + {b^3} - 3abx = 0$ we get $q = {a^3} + {b^3}$ and $p = - 3ab \implies - \frac{{{p^3}}}{{27}} = {a^3}{b^3}$. Now note that, $${a^6} - \left( {{a^3} + {b^3}} \right) \cdot {a^3} + {a^3}{b^3} = 0.$$ Taking ${a^3} = z$ we get, $${z^2} - qz - \frac{{{p^3}}}{{27}} = 0 \implies z = \frac{{q \pm \sqrt {{q^2} + \frac{{4{p^3}}}{{27}}} }}{2} \implies a = \sqrt[3]{{\frac{q}{2} + \sqrt {\frac{{{q^2}}}{4} + \frac{{{p^3}}}{{27}}} }}.$$ Hence, $b = \sqrt[3]{{\frac{q}{2} - \sqrt {\frac{{{q^2}}}{4} + \frac{{{p^3}}}{{27}}} }}$. Now, \begin{align*} &{x^3} + {a^3} + {b^3} - 3abx = 0\\ \implies & \left( {x + a + b} \right)\left( {{x^2} + {a^2} + {b^2} - ax - bx - ab} \right) = 0\\ \implies & x + a + b = 0\quad {\text{or}}\quad {x^2} - \left( {a + b} \right)x + \left( {{a^2} + {b^2} - ab} \right) = 0\\ \implies & x = - \left( {a + b} \right)\quad {\text{or}}\quad x = \frac{{a + b \pm \left( {a - b} \right)\sqrt 3 i}}{2}. \end{align*} Now, putting the values of $a$ and $b$, we will get the solutions.

Now we find the solutions of ${x^3} + p{x^2} + qx + r = 0$.

Putting $x=m+n$ in the equation we get, \begin{align*} &{\left( {m + n} \right)^3} + p{\left( {m + n} \right)^2} + q\left( {m + n} \right) + r = 0\\ \implies & {m^3} + \left( {3n + p} \right){m^2} + \left( {3{n^2} + 2pn + q} \right)m + \left( {{n^3} + p{n^2} + qn + r} \right) = 0. \end{align*}

Now letting $3n+p=0\implies n=-p/3$. Now putting the value of $n$ in the equation we get, $${m^3} + \left( {q - \frac{{{p^2}}}{3}} \right)m + \left( {\frac{{2{p^3}}}{{27}} - \frac{{pq}}{3} + r} \right) = 0.$$

Now the problem reduced to first problem.