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.
 A: 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.
A: With free computer algebra system Maxima 5.42.1

