Solving $8x^3 - 6x + 1$ using Cardano's method

Solve for the first root of $$8x^3 - 6x + 1 = 0$$

After solving I get $$\sqrt[3]{\frac{-1 + \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$$, which is not a solution to the cubic equation: Here's how I come up with:

Using Cardano's method:
let $$x = y - \frac{b}{3a}$$

Then compress:
Since $$b$$ is $$0$$, then it turns out to be $$8y^3 - 6y + 1 = 0$$. Divide $$8$$ to both sides.
$$y^3 - \frac{3}{4}y + \frac{1}{8} = 0$$

Let $$3st = \frac{-3}{4}$$ and $$s^3 - t^3 = \frac{-1}{8}$$
Now: \begin{align} \left(\frac{-1}{4t}\right)^{3} - t^{3} &= \frac{1}{8}\\ ...\\ 8t^6 - t^3 + \frac{1}{8} &= 0\\ \end{align}

I'll uncompress the equation above so it becomes quadratic: \begin{align} 8t^2 - t + \frac{1}{8} &= 0\\ ...\\ \left(\frac{1 + \sqrt{3}i}{16}\right)\left(\frac{1 - \sqrt{3}i}{16}\right)\\ \end{align}

Then take the cuberoot to get $$t$$ (only taking the positive root): $$\sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$$

Since $$x = y - \frac{0}{24}$$, which is similar to $$x = y$$ and $$y = s - t$$, then:

$$t = \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$$ $$s = \sqrt[3]{\frac{1 - \sqrt{3}i}{16}}$$ \begin{align} y &= s - t\\ y &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\ x &= y + 0\\ x &= \sqrt[3]{\frac{1 - \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}\\ \end{align}

Checking if it is a solution, it turns out that it is not. Note: I'm new to this method so its unclear to me why it dont work and PLEASE dont mark this as a duplicate. Thanks.

• Can't be sure but ... What values of those cube roots did you use? You have to remember that the cube root of a complex number has three possible values. And (THIS IS IMPORTANT) you must select those cube roots $s$ and $t$ in such a way that they satisfy the equation $3st=-3/4$. Some calculator of cube roots may only output the one with smallest positive argument, and that will often fail to observe this condition. – Jyrki Lahtonen Nov 16 '18 at 8:05
• To test whether may hunch is correct please do the following. Let $t$ be whatever your calculator gives it. THEN LET $s$ be a solution of $3st=-3/4$. So let $s=-1/(4t)$. AND ONLY THEN CALCULATE $s-t$. – Jyrki Lahtonen Nov 16 '18 at 8:07
• It is possible that something else went wrong, but that is my first guess, because other askers have failed at the exact same spot before (meaning that your question may be a duplicate). – Jyrki Lahtonen Nov 16 '18 at 8:08
• In other words, I suspect that your question is a duplicate of this. I'm actually mildly surprised, if that is the earliest incarnation of that problem on our site. – Jyrki Lahtonen Nov 16 '18 at 8:10
• Since Im new to Cardano's method, I did not know that complex numbers have also nth roots, which has 3 possible roots. I thought its easy to represent it as $\sqrt[3]{a + bi}$. – MMJM Nov 16 '18 at 8:17

You may have a sign error. I render $$s^3$$ as $$-(1-i\sqrt{3})/16$$ but you seem to have $$+(1-i\sqrt{3})/16$$. But even with the sign correction (which probably does get you a good answer), your approach is not best. See below.
The solution is easy. When you get a root for $$t$$, do not solve independently for $$s$$. Use the fact that $$st=-(1/4), s=-(1/4t)$$ to get an expression for $$s$$ that has the same cube root radical as the one for $$t$$. Now your intended difference $$s-t$$ contains only a single cube root radical, and its three possible values correspond properly to the three roots of the cubic equation.
• I got $$\frac{-1}{4\sqrt[3]{\frac{1+\sqrt{3}i}{16}}} - \sqrt[3]{\frac{1+\sqrt{3}i}{16}}$$. Thanks!! – MMJM Nov 17 '18 at 1:27