By using properties of determinants, show that:

(i) $\Delta=\left|\begin{array}{ccr}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have:

$\Delta=\left|\begin{array}{lll}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$

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

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

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

$=(a+b+c)^{3}\left|\begin{array}{lll}1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1\end{array}\right|$

Expanding along C3, we have:

$\Delta=(a+b+c)^{3}(-1)(-1)=(a+b+c)^{3}$

Hence, the given result is proved.

(ii) $\Delta=\left|\begin{array}{clr}x+y+2 z & x & y \\ z & y+z+2 x & y \\ z & x & z+x+2 y\end{array}\right|$

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we have:

$\begin{aligned} \Delta &=\left|\begin{array}{lll}2(x+y+z) & x & y \\ 2(x+y+z) & \quad y+z+2 x & y \\ 2(x+y+z) & x & z+x+2 y\end{array}\right| \\ &=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 1 & y+z+2 x & y \\ 1 & x & z+x+2 y\end{array}\right| \end{aligned}$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:

$\begin{aligned} \Delta &=2(x+y+z)\left|\begin{array}{llll}1 & & x & y \\ 0 & & x+y+z & 0 \\ 0 & & 0 & x+y+z\end{array}\right| \\ &=2(x+y+z)\left|\begin{array}{lll}1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right| \end{aligned}$

Expanding along R3, we have:

$\Delta=2(x+y+z)^{3}(1)(1-0)=2(x+y+z)^{3}$

Hence, the given result is proved.