『設 p, q 是實數，且 a,b,c 是 x^3+px+q=0 的三根，則判別式=[(a-b)(b-c)(c-a)]^2=-4p^3-27q^2』
嘗試用凡德夢行列式，終於做出結論，要用到det(AB)=det(A)*det(B)
\({{\left( a-b \right)}^{2}}{{\left( b-c \right)}^{2}}{{\left( c-a \right)}^{2}}\)
=\(det
\left (\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2
\end{array} \right) *
det
\left(\begin{array}{ccc}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right)
\)
=\(det
\left(\begin{array}{ccc}
3 & s_1 & s_2 \\
s_1 & s_2 & s_3 \\
s_2 & s_3 & s_4
\end{array}\right)
\)
=\(det
\left(\begin{array}{ccc}
3 & 0 & -2p \\
0 & -2p & -3q \\
-2p & -3q & 2p^2
\end{array}\right) \)
\(Hence, result. \)
其中\(s_1=a+b+c, \;s_2=a^2+b^2+c^2,\; s_3=a^3+b^3+c^3,\; s_4=a^4+b^4+c^4 \)
橢圓兄的招都頗巧
