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 $\left.C_2 \rightarrow C_2-C_3\right]$

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

We have to prove,

$\left|\begin{array}{ccc}y^2 z^2 & y z & y+z \\ z^2 x^2 & z x & z+x \\ x^2 y^2 & x y & x+y\end{array}\right|=0$

$$\begin{array}{r} \therefore\quad \mathrm{LHS}=\left|\begin{array}{rll} y^2 z^2 & y z & y+z \\ z^2 x^2 & z x & z+x \\ x^2 y^2 & x y & x+y \end{array}\right|=\frac{1}{x y z}\left|\begin{array}{rrr} x y^2 z^2 & x y z & x y+x z \\ x^2 y z^2 & x y z & y z+x y \\ x^2 y^2 z & x y z & x z+y z \end{array}\right| \\ {\left[\because R_1 \rightarrow x R_1, R_2 \rightarrow y R_2, R_3 \rightarrow z R_3\right]} \end{array}$$

$$=\frac{1}{x y z}(x y z)^2\left|\begin{array}{ccc} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right|\quad$$ [taking $(x y z)$ common from $\mathrm{C}_1$ and $\mathrm{C}_2$]

$$\begin{aligned} & =x y z\left|\begin{array}{ccc} y z & 1 & x y+y z+z x \\ x z & 1 & x y+y z+z x \\ x y & 1 & x y+y z+z x \end{array}\right|\left[C_3 \rightarrow C_3+C_1\right] \\ & =x y z(x y+y z+z x)\left|\begin{array}{ccc} y z & 1 & 1 \\ x z & 1 & 1 \\ x y & 1 & 1 \end{array}\right| \end{aligned}$$

$$\begin{array}{lr} =0 & {\left[\text { taking }(x y+y z+z x) \text { common from } C_3\right]} \\ =\text { RHS } & {\left[\text { since, } C_2 \text { and } C_3 \text { are identicals }\right]} \\ \text { Hence proved. } \end{array}$$

We have to prove,

$\left|\begin{array}{ccc}y+z & z & y \\ z & z+x & x \\ y & x & x+y\end{array}\right|=4 x y z$

$$\begin{aligned} \mathrm{LHS} & =\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right| \\ & =\left|\begin{array}{ccc} y+z+z+y & z & y \\ z+z+x+x & z+x & x \\ y+x+x+y & x & x+y \end{array}\right| \quad\left[\because C_1 \rightarrow C_1+C_2+C_3\right] \end{aligned}$$

$$=2\left|\begin{array}{ccc} (y+z) & z & y \\ (z+x) & z+x & x \\ (x+y) & x & x+y \end{array}\right|\quad$$ [taking 2 common from $\mathrm{C}_1$ ]

$$\begin{array}{ll} =2\left|\begin{array}{ccc} y & z & y \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| & {\left[\because C_1 \rightarrow C_1-C_2\right]} \\ =2\left|\begin{array}{ccc} 0 & z-x & -x \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| & {\left[\because R_1 \rightarrow R_1-R_3\right]} \end{array}$$

$$\begin{aligned} &\begin{aligned} & =2\left[y\left(x z-x^2+x z+x^2\right)\right] \\ & =4 x y z=\mathrm{RHS} \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

We have to prove,

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

$\therefore \quad$ LHS $=\left|\begin{array}{ccc}a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

$$\begin{aligned} =\left|\begin{array}{ccc} a^2+2 a-2 a-1 & 2 a+1-a-2 & 0 \\ 2 a+1-3 & a+2-3 & 0 \\ 3 & 3 & 1 \end{array}\right| & \\ & {\left[\because R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3\right] } \end{aligned}$$

$$\begin{aligned} &=\left|\begin{array}{ccc} (a-1)(a+1) & (a-1) & 0 \\ 2(a-1) & (a-1) & 0 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^2\left|\begin{array}{ccc} (a+1) & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right| \\ &\text { [taking } \left.(a-1) \text { common from } R_1 \text { and } R_2 \text { each] }\right] \end{aligned}$$

$$\begin{aligned} & =(a-1)^2[1(a+1)-2]=(a-1)^3 \\ & =\text { RHS }\quad\text{Hence proved.} \end{aligned}$$