We have, $\left|\begin{array}{ccc}a+x & y & z \\ x & a+y & z \\ x & y & a+z\end{array}\right|=\left|\begin{array}{ccc}a & -a & 0 \\ 0 & a & -a \\ x & y & a+z\end{array}\right|$

$$\left[\begin{array}{cc} \because & R_1 \rightarrow R_1-R_2 \\ \text { and } & R_2 \rightarrow R_2-R_3 \end{array}\right]$$

$$\begin{aligned} & =\left|\begin{array}{ccc} a & 0 & 0 \\ 0 & a & -a \\ x & x+y & a+z \end{array}\right| \quad \left[\because C_2 \rightarrow C_2+C_1\right]\\ & =a\left(a^2+a z+a x+a y\right) \\ & =a^2(a+z+x+y) \end{aligned}$$

We have, $\quad\left|\begin{array}{ccc}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{array}\right|=x^2 y^2 z^2\left|\begin{array}{ccc}0 & x & x \\ y & 0 & y \\ z & z & 0\end{array}\right|$

$$\begin{aligned} &\text { [taking } x^2, y^2 \text { and } z^2 \text { common from } C_1, C_2 \text { and } C_3 \text {, respectively] }\\ &\begin{aligned} & =x^2 y^2 z^2\left|\begin{array}{ccc} 0 & 0 & x \\ y & -y & y \\ z & z & 0 \end{array}\right| \\ & =x^2 y^2 z^2[x(y z+y z)] \\ & =x^2 y^2 z^2 \cdot 2 x y z=2 x^3 y^3 z^3 \end{aligned} \quad\left[\because C_2 \rightarrow C_2-C_3\right] \end{aligned}$$

We have, $\quad\left|\begin{array}{ccc}3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z\end{array}\right|$

Applying, $C_1 \rightarrow C_1+C_2+C_3$,

$$\begin{aligned} & =\left|\begin{array}{ccc} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z \end{array}\right| \\ & =(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{array}\right| \quad \text { [taking } \left.(x+y+z) \text { common from column } C_1\right]\\ & =(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 0 & 2 y+x & x-y \\ 0 & x-z & 2 z+x \end{array}\right| \quad {\left[\because R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1\right]}\\ \end{aligned}$$

$$\begin{aligned} &\text { Now, expanding along first column, we get }\\ &\begin{aligned} & (x+y+z) \cdot 1[(2 y+x)(2 z+x)-(x-y)(x-z)] \\ = & (x+y+z)\left(4 y z+2 y x+2 x z+x^2-x^2+x z+y x-y z\right) \\ = & (x+y+z)(3 y z+3 y x+3 x z) \\ = & 3(x+y+z)(y z+y x+x z) \end{aligned} \end{aligned}$$

We have, $\left|\begin{array}{ccc}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4\end{array}\right|=\left|\begin{array}{ccc}2 x+4 & 2 x+4 & 2 x \\ x & x+4 & x \\ x & x & x+4\end{array}\right|\quad$ $$\left[\because R_1 \rightarrow R_1+R_2\right]$$

$$\begin{aligned} &=\left|\begin{array}{ccc} 2 x & 2 x & 2 x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|+\left|\begin{array}{ccc} 4 & 4 & 0 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|\\ &\text { [here, given determinant is expressed in sum of two determinants] }\\ &=2 x\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|+4\left|\begin{array}{ccc} 1 & 1 & 0 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right| \end{aligned}$$

[taking $2 x$ common from first row of first determinant and 4 from first row of second determinant]

Applying $C_1 \rightarrow C_1-C_3$ and $C_2 \rightarrow C_2-C_3$ in first and applying $C_1 \rightarrow C_1-C_2$ in second, we get

$=2 x\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 4 & x \\ -4 & -4 & x+4\end{array}\right|+4\left|\begin{array}{ccc}0 & 1 & 0 \\ -4 & x+4 & x \\ 0 & x & x+4\end{array}\right|$

$$\begin{aligned} &\text { Expanding both the along first column, we get }\\ &\begin{aligned} 2 x & {[-4(-4)]+4[4(x+4-0)] } \\ & =2 x \times 16+16(x+4) \\ & =32 x+16 x+64 \\ & =16(3 x+4) \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { We have, } \quad &\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \\ &=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\quad \left[\because R_1 \rightarrow R_1+R_2+R_3\right] \end{aligned}$$

$$\begin{gathered} =(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \\ {[\text { taking }(a+b+c) \text { common from the first row }]} \end{gathered}$$

$$=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -(a+b+c) & 2 b \\ (a+b+c) & (a+b+c) & (c-a-b) \end{array}\right|$$

$\left[\because C_1 \rightarrow C_1-C_3\right.$ and $\left.C_2 \rightarrow C_2-C_3\right]$

$$\begin{aligned} &\text { Expanding along } R_1 \text {, }\\ &\begin{aligned} & =(a+b+c)\left[1\left\{0+\left(a+b+c^2\right\}\right]\right. \\ & =(a+b+c)\left[(a+b+c)^2\right] \\ & =(a+b+c)^3 \end{aligned} \end{aligned}$$