If each element of a determinant of third order with value $A$ is multiplied by $3$,then the value of the newly formed determinant is: (in $A$)

By using properties of determinants,show that:
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

If ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$,then the value of $\sum\limits_{r = 1}^n {D_r} = $

Let $D_{k} = \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix}$. If $\sum_{k=1}^{n} D_{k} = 96$,then $n$ is equal to

By using properties of determinants,show that:
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

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