If $a, b, c > 0$ and $x, y, z \in R$,then the value of the determinant $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^y + b^{-y})^2 & (b^y - b^{-y})^2 & 1 \\ (c^z + c^{-z})^2 & (c^z - c^{-z})^2 & 1 \end{array} \right|$ is equal to:

If ${f_n}(x)$,${g_n}(x)$,${h_n}(x)$ for $n = 1, 2, 3$ are polynomials in $x$ such that ${f_n}(a) = {g_n}(a) = {h_n}(a)$ for $n = 1, 2, 3$,then the determinant $F(x) = \left| \begin{matrix} {f_1}(x) & {f_2}(x) & {f_3}(x) \\ {g_1}(x) & {g_2}(x) & {g_3}(x) \\ {h_1}(x) & {h_2}(x) & {h_3}(x) \end{matrix} \right|$ at $x = a$ is equal to:

Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is:

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

If $a \neq b \neq c$, $\Delta_1=\left|\begin{array}{lll}1 & a^2 & b c \\ 1 & b^2 & c a \\ 1 & c^2 & a b\end{array}\right|$, $\Delta_2=\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$ and $\frac{\Delta_1}{\Delta_2}=\frac{6}{11}$, then $11(a+b+c)=$

$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right| = $