If $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$, then $k = $

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

By using properties of determinants, show that:

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

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to ...... .

If $f(x) = \left| {\begin{array}{*{20}{c}}{x - 3}&{2{x^2} - 18}&{3{x^3} - 81}\\{x - 5}&{2{x^2} - 50}&{4{x^3} - 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.