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:
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

For non-zero $a, b, c$,if $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$,then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

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

Let $a, b, c$ be such that $b + c \ne 0$. If $\left| \begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array} \right| + \left| \begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} \cdot a & (-1)^{n+1} \cdot b & (-1)^n \cdot c \end{array} \right| = 0$,then $n$ is equal to: