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Sa se calculeze determinantul matricei A=\( \begin{pmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
{x_1}^2 & {x_2}^2 & {x_3}^2
\end{pmatrix} \ \), unde \( x_1, x_2, x_3 \in \mathbb{R} \) sunt radacinile ecuatiei \( x^3+px+q=0 \).

\( S_1=s_1=x_1+x_2+x_3=0 \)
\( S_2=x_1x_2+x_2x_3+x_3x_1=p \)
\( S_3=x_1x_2x_3=-q \)

\( s_2=\sum x^2=S_1^2-2S_2=-2p \)
\( s_3=\sum x^3=S_1 \cdot S_2 + 3S_3=-3q \)
\( s_4=\sum x^4=s_2^2-(2S_2^2-4S_3 \cdot S_1)=2p^2 \)

\( A \cdot ^t A= \)\( \begin{pmatrix} 3 & s_1 & s_2 \\ s_1 & s_2 & s_3 \\ s_2 & s_3 & s_4 \end{pmatrix}\ \)

\( det{A \cdot ^tA}=3(s_2s_4-s_3^2)-s1(s_1s_4-s_2s_3)+s_2(s1s3-s2^2)=-27q^2-4p^3 \)

\( detA=\sqrt{-27q^2-4p^3} \)