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.

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Nov 16 '18 at 8:05
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    $\begingroup$ 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 $s-t$. $\endgroup$ – Jyrki Lahtonen Nov 16 '18 at 8:07
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    $\begingroup$ 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). $\endgroup$ – Jyrki Lahtonen Nov 16 '18 at 8:08
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Nov 16 '18 at 8:10
  • $\begingroup$ 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 $\sqrt[3]{a + bi}$. $\endgroup$ – 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.

When you have an expression with two different complex cube roots there are nine choices for this combination. Any of three roots or the first cube root could be paired with any of three cube roots for the second. Yet, of course, only three of the combinations can be correct for the cubic equation you started with. If your calculator/computer program does not choose the "principal values" of the cube roots in the right way for this particular equation, you go wrong (even without other errors).

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.

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    $\begingroup$ I got $$ \frac{-1}{4\sqrt[3]{\frac{1+\sqrt{3}i}{16}}} - \sqrt[3]{\frac{1+\sqrt{3}i}{16}}$$. Thanks!! $\endgroup$ – MMJM Nov 17 '18 at 1:27

With free computer algebra system Maxima 5.42.1 enter image description here


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