1 | initial version |

First of all, let us move the fraction $1/6$ from the second formula inside the square root. As $1/36$. Then we get inside $\sqrt[3]{\ \dots\ }$ the following expression. $$ -\frac g2 +\sqrt{\left(\frac g2\right)^2 +\left(\frac f3\right)^3}\ . $$ This is the way i would write the formula.

The second formula is correct.

The first one cannot work.

If the question is how to write `sage`

code that checks that two symbolic algebraic expressions
involving radicals inside radicals are, or are not equal - a purely programatic question, then this post is not an answer.
If the way the formulas arise is the target, and if the check is / should be combined with the mathematic knowledge, then
the following is what i would prefer.
(There is also a detail in the choice of the third root in the two terms appearing in the formulas. They have to be correlated. This correlation cannot be explained in a trivial manner to the sage compiler by simply using a code fragment of the shape
`( ... )^(1/3) + ( ... )^(1/3)`

, except we tacitly understand that the product of the cubic roots chosen ( ... )^(1/3)( ... )^(1/3)
is fixed.
And this is the reason why some textbooks prefer to put the second cubic root in a cumbersome expression.)

First of all, let $\epsilon$ be a primitive third root of unity. We can then factor the expression $$ X^3 + A^3 + B^3 - 3XAB $$ as follows (over $\mathbb Q[A,B;X]$)

```
var( 'X,A,B' );
factor( X^3 + A^3 + B^3 - 3*A*B*X )
```

getting

```
(A^2 - A*B + B^2 - A*X - B*X + X^2)*(A + B + X)
```

-- and this is enough for our strict purposes, the factor $(X+A+B)$ is enough for the following -- but we can even ``force'' three linear factors after extending the coefficients from $\mathbb Q$ to $\mathbb Q[\epsilon]=\mathbb Q[\sqrt{-3}]$ . The corresponding code is:

```
F.<e> = CyclotomicField( 3 )
R.<A,B,X> = PolynomialRing( F )
factor( X^3 + A^3 + B^3 - 3*A*B*X )
```

And the factors are coming with the expected Galois symmetry:

```
(A + B + X) * (A + (e)*B + (-e - 1)*X) * (A + (-e - 1)*B + (e)*X)
```

Humanly written:
$$ X^3 + A^3 + B^3 - 3XAB = (X+A+B)(X^2+A^2 +B^2-XA-XB-AB) \ .$$
Even better:
$$ X^3 + A^3 + B^3 - 3XAB= (X+A+B)(X+\epsilon A+\epsilon^2 B)(X+\epsilon^2 A+\epsilon B)\ .
$$
Now we plug in the following symbolic values for $A$ and $B$.
Since i hate an excessive use of denominators (in $\LaTeX$), let $G=g/2$ and $F=f/3$. Let $A,B$ be the two values
$$ A ,B = -\left(\ - G\pm \sqrt{F^3+G^2}\ \right)^{1/3}\ .$$
*Tacit* convention:
The two cubic roots are taken such that $AB$, constrained to be
$$ AB = \left(- G+\sqrt{F^3+G^2}\right)^{1/3}\left(- G-\sqrt{F^3+G^2}\right)^{1/3}$$
$$\qquad = \left(G^2 -(F^3+G^2)\right)^{1/3} = -(F^3)^{1/3}\in{F,\epsilon F,\epsilon^2 F}\ , $$
is *de facto* exactly $AB=-F$.
Then
$$ X^3 + A^3 + B^3 - 3XAB = X^3 -\left(- G+\sqrt{F^3+G^2}\right)-\left(- G-\sqrt{F^3+G^2}\right) -3X(-F)=\dots
$$
This is $X^3 +3XF+2G = X^3+fX+g$.

The factor $(X+A+B)$ of $X^3+A^3+B^3-3ABX$ exposes the root $-(A+B)$ as in the second posted formula, the correct one. Let us give in `sage`

a particular example:

```
sage: ( x^3 + 6*x + 2 ).roots( multiplicities=0 )
[1/2*2^(2/3)*(-I*sqrt(3) + 1) - 1/2*2^(1/3)*(I*sqrt(3) + 1),
1/2*2^(2/3)*(I*sqrt(3) + 1) - 1/2*2^(1/3)*(-I*sqrt(3) + 1),
-2^(2/3) + 2^(1/3)]
```

Here, $f=6$, $F=f/3=2$, $g=2$, $G=g/2=1$, then $\sqrt{F^3+G^2}=\sqrt{2^3+1^2}=3$, and we build $-1\pm 3$ and so on.

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