<s>[INST] How to find the value of $\sqrt{\Delta}$ for the given question? If it is given that $\alpha , \beta, \gamma$ are the roots of $x^3+x^2-2x+1=0$, then find the value $\sqrt{\Delta}$, where $$
\Delta=\left|\begin{array}{ccc}
\gamma^2 & \beta^2 & 2 \beta \gamma-\alpha^2 \\
2 \alpha \gamma-\beta^2 & \alpha^2 & \gamma^2 \\
\alpha^2 & 2 \alpha \beta-\gamma^2 & \beta^2
\end{array}\right|
$$
I tried expanding the determinant and it comes out to be,
\begin{align*} 
 \Delta &= \alpha^{6} - 6 \alpha^{4} \beta \gamma + 2 \alpha^{3} \beta^{3} + 2 \alpha^{3} \gamma^{3} + 9 \alpha^{2} \beta^{2} \gamma^{2} - 6 \alpha \beta^{4} \gamma - 6 \alpha \beta \gamma^{4} + \beta^{6} + 2 \beta^{3} \gamma^{3} + \gamma^{6} \\ 
 &=  (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)^2
\end{align*}
But I don't know how to proceed from here, also I believe that there should be another way to do it, as this expansion is way too tedious to do. Any help is appreciated.
Edit: I managed to get the answer $\sqrt{-7}$ using Newton's identities.
 [/INST] By confort put $a,b,c$ for the three roots. You do have
$$a+b+c=-1\\ab+ac+bc=-2\\abc=-1\tag1$$ Besides $$(a+b+c)^3=a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc\\(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\tag2$$ From $(2)$ and $(1)$
$$-1=a^3+b^3+c^3+3a^2(-1-a)+3b^2(-1-b)+3c^2(-1-c)-6$$ then $$2(a^3+b^3+c^3)=-3(a^2+b^2+c^2)-5=-3((-1)^2+4)-5=-20$$ Finally, using your
$\Delta=(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)^2$
we get
$$\Delta=(a^3+b^3+c^3-3abc)^2=(-10+3)^2=49$$
</s>