The value of the determinant $\left| \begin{array}{ccc} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{array} \right|$ is

$\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$ is not equal to

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

If $\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ a & b & c \\ l & m & n\end{array}\right|=(-1)^K\left|\begin{array}{ccc}m & n & l \\ b & c & a \\ \beta & \gamma & \alpha\end{array}\right|$,then the least value of $K$ is

Evaluate the determinant: $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{array} \right|$

Without expanding the determinant,prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$