Does there exists another approach to solve for the product of such expression? Problem: if the real roots of $x^3-3x+1$ are $\alpha , \beta $ and $\gamma,$ then what is the value of cyclic $(\alpha^2-\gamma)\;?$
Here is my approach, using trigonometry. Is my work correct?

 A: The problem can be brute-forced algebraically, but the calculations involve polynomial resultants and are laborious to do by hand, though easily computed using a CAS.
Let $\,x_1=\alpha, x_2=\beta, x_3=\gamma\,$, then the polynomial in $\,y\,$ with roots $\,x_i^2\,$ is:
$$\text{res}(x^3-3x+1, y-x^2, x)=y^3 - 6 y^2 + 9 y - 1 \tag{1}$$
The polynomial in $\,z,y\,$ with roots $\,x_i^2-x_j = y_i-x_j\,$ is :
$$
\text{res}(x^3-3x+1, z-y+x, x) = -y^3 + 3 y^2 z - 3 y z^2 + 3 y + z^3 - 3 z - 1 \tag{2}
$$
Then, the polynomial in $\,z\,$ alone is found by eliminating $\,y\,$ between $(1)$ and $(2)$, which gives:
$$
\begin{align}
& \text{res}(y^3 - 6 y^2 + 9 y - 1, -y^3 + 3 y^2 z - 3 y z^2 + 3 y + z^3 - 3 z - 1, y)
\\ =\, &(z - 2)^3 (z^3 - 6 z^2 + 3 z + 19) (z^3 - 6 z^2 + 3 z + 1) \tag{3}
\end{align}
$$
Three of the $\,z\,$ roots are $\,x_i^2-x_i\,$, which are the roots of:
$$
\text{res}(x^3-3x+1, z-x^2+x, x) = z^3 - 6 z^2 + 3 z + 1 \tag{4}
$$
This corresponds to the last factor in $\,(3)\,$, which leaves the two possibilities $(\dagger)\,$:

*

*$(z-2)^3 = 0\,$ with the triple root $\,2\,$ so $\,z_1z_2z_3=8\,$, which is OP's solution;


*$z^3 - 6 z^2 + 3 z + 19\,$ with the product of the roots $\,z_1z_2z_3=-19\,$, which is the second solution linked in my comment.
$(\dagger)\;$ The relevant sextic in $\,(3)\,$ can be factored into two cubics in $\,\binom{6}{3}\,$ ways, but we know that $\,(x_1^2-x_2)+(x_2^2-x_3)+(x_3^2-x_1)=6\,$ and the only factorization where the coefficient of $\,x^2\,$ is $\,-6\,$ in a cubic factor is where the factors are chosen to be $\,(z-2)^3\,$ and $\,z^3 - 6 z^2 + 3 z + 19\,$.
