ここで勉強すれば数学検定１級の壁は超えられるか。

$\begin{bmatrix} \begin{pmatrix} a \\ 1 \end{pmatrix} & \begin{pmatrix} a \\ 2 \end{pmatrix} & \begin{pmatrix} a \\ 3 \end{pmatrix} \\ \begin{pmatrix} b \\ 1 \end{pmatrix} & \begin{pmatrix} b \\ 2 \end{pmatrix} & \begin{pmatrix} b \\ 3 \end{pmatrix} \\ \begin{pmatrix} c \\ 1 \end{pmatrix} & \begin{pmatrix} c \\ 2 \end{pmatrix} & \begin{pmatrix} c \\ 3 \end{pmatrix} \end{bmatrix}$

$=\begin{vmatrix} a & \dfrac {a\left( a-1\right) }{2} & \dfrac {a\left( a-1\right) \left( a-2\right) }{6} \\ b & \dfrac {b\left( b-1\right) }{2} & \dfrac {b\left( b-1\right) \left( b-2\right) }{6} \\ c & \dfrac {c\left( c-1\right) }{2} & \dfrac {c\left( c-1\right) \left( c-2\right) }{6} \end{vmatrix}$

$=abc\begin{vmatrix} 1 & \dfrac {a-1}{2} & \dfrac {\left( a-1\right) \left( a-2\right) }{6} \\ 1 & \dfrac {b-1}{2} & \dfrac {\left( b-1\right) \left( b-2\right) }{6} \\ 1 & \dfrac {c-1}{2} & \dfrac {\left( c-1\right) \left( c-2\right) }{6} \end{vmatrix}$

$=\dfrac {abc}{12}\begin{vmatrix} 1 & a-1 & a^{2}-3a+2 \\ 0 & b-a & \left( b^{2}-a^{2}\right) -3\left( b-a\right) \\ 0 & c-a & c^{2}-a^{2}-3\left( c-a\right) \end{vmatrix}$

$=\dfrac {abc}{12}\begin{vmatrix} b-a & \left( b-a\right) \left(b+a-3\right) \\ c-a & \left( c-a\right) \left(c+a-3\right) \end{vmatrix}$

$=\dfrac {abc}{12}\left( b-a\right) \left( c-a\right) \left\{ \left( c+a-3\right) -\left( b+a-3\right) \right\}$

$=\dfrac {abc}{12}\left( a-b\right) \left( b-c\right) \left( c-a\right)$・・・答