Let $D_1 =$ $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\c&d&{c + d}\\a&b&{a - b}\end{array}\,} \right|$ and $D_2 =$ $\left| {\,\begin{array}{*{20}{c}}a&c&{a + c}\\b&d&{b + d}\\a&c&{a + b + c}\end{array}\,} \right|$ then the value of $\frac{{{D_1}}}{{{D_2}}}$ where $b \ne 0$ and $ad \ne bc$, is

The value of $\sum\limits_{n = 1}^N {{U_n},} $ if ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ is

If $\mathrm{a, b, c},$ are in $\mathrm{A.P}$, then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$ is

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right|$ is equal to 

If ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$ such that $\left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right| = \lambda $, then the value of $\lambda $is

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x - 1)}&{(x + 1)x}\\{3x(x - 1)}&{x(x - 1)(x - 2)}&{(x + 1)x(x - 1)}\end{array}} \right|$ then $f(100)$ is equal to