$\left| {\begin{array}{ccc} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right| = $

Without expanding,prove that $\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = 0$.

If $\Delta = \begin{vmatrix} a + x & b & c \\ b & x + c & a \\ c & a & x + b \end{vmatrix}$,which of the following is a factor for the above determinant?

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}, \dots$ are in $G.P.$,then the value of the determinant $\left| \begin{array}{ccc} \log {a_n} & \log {a_{n+1}} & \log {a_{n+2}} \\ \log {a_{n+3}} & \log {a_{n+4}} & \log {a_{n+5}} \\ \log {a_{n+6}} & \log {a_{n+7}} & \log {a_{n+8}} \end{array} \right|$ is

If $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x \neq y \neq z$, then $1+x y z$ is equal to

Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$