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

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $.....$

If $a, b$ and $c$ are real numbers, and

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

Show that either $a+b+c=0$ or $a=b=c$.

By using properties of determinants, show that:

$\left|\begin{array}{ccc}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|=k^{2}(3 x+k)$

$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

Which of the following values of $\alpha$ satisfy the equation

$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?

$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$