If $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then the value of $abc(ab + bc + ca)$ is

If $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{matrix} \right| = 5$,then $\left| \begin{matrix} bc^2 - b^2c & a^2c - ac^2 & ab^2 - ba^2 \\ b^2 - c^2 & c^2 - a^2 & a^2 - b^2 \\ c - b & a - c & b - a \end{matrix} \right|$ is equal to

If $x, y, z \in \mathbb{R}$,then the value of the determinant $\left|\begin{array}{lll}\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|$ is:

If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to

By using properties of determinants,show that:
$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3y+k)$

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