If $\omega$ is an imaginary cube root of unity, then the value of the determinant $\left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$ is

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|=$

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}, \dots$ are in $G$.$P$. and ${a_i} > 0$ for each $i$,then the value of the determinant $\Delta = \begin{vmatrix} \log {a_n} & \log {a_{n+2}} & \log {a_{n+4}} \\ \log {a_{n+6}} & \log {a_{n+8}} & \log {a_{n+10}} \\ \log {a_{n+12}} & \log {a_{n+14}} & \log {a_{n+16}} \end{vmatrix}$ is equal to

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} \right|$ is equal to:

Show that $\Delta=\left|\begin{array}{ccc} (y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & y z & (x+y)^{2} \end{array}\right|=2 x y z(x+y+z)^{3}$