<s>[INST] 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.
 [/INST] With free computer algebra system Maxima 5.42.1

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