Solve the following determinant equations:
(i) $\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0$
(ii) $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$
(iii) $\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8\end{array}\right|=0$
(iv) $\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2}\end{array}\right|=0, a \neq b$
(v) $\left|\begin{array}{ccc}x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4\end{array}\right|=0$
(vi) $\left|\begin{array}{lll}1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3}\end{array}\right|=0, b \neq c$'
(vii) $\left|\begin{array}{ccc}15-2 x & 11-3 x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13\end{array}\right|=0$
(viii) $\left|\begin{array}{ccc}1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2\end{array}\right|=0$
(ix) $\left|\begin{array}{ccc}3 & -2 & \sin (3 \theta) \\ -7 & 8 & \cos (2 \theta) \\ -11 & 14 & 2\end{array}\right|=0$
(x) $\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4+x\end{array}\right|=0$
(x) Given: $\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4+x\end{array}\right|=0$
LHS $=\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4+x\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1}$
$=\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 4+x-4+x & 4-x-4-x & 4+x-4-x \\ 4+x & 4+x & 4+x\end{array}\right|$
$=\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 2 x & -2 x & 0 \\ 4+x & 4+x & 4+x\end{array}\right|$
Taking $(2 x)$ common from $R_{2}$
$=(2 x)\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 1 & -1 & 0 \\ 4+x & 4+x & 4+x\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}-R_{1}$
$=(2 x)\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 1 & -1 & 0 \\ 4+x-4+x & 4+x-4-x & 4+x-4-x\end{array}\right|$
$=(2 x)\left|\begin{array}{ccc}4-x & 4+x & 4+x \\ 1 & -1 & 0 \\ 2 x & 0 & 0\end{array}\right|$
Expanding through $C_{3}$
$=(2 x)[(4+x)(1 \times 0+1 \times 2 x)]$
$=(2 x)[(4+x)(2 x)]$
$=(2 x)^{2}(4+x)$
Thus, $\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4+x\end{array}\right|=(2 x)^{2}(4+x)$
But it is given that, $\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4+x\end{array}\right|=0$
$\Rightarrow(2 x)^{2}(4+x)=0$
$\Rightarrow(2 x)^{2}=0$ or $(4+x)=0$
$\Rightarrow x=0$ or $x=-4$
Hence, $x=0,-4$.