By using properties of determinants, show that:

(i) $\Delta=\left|\begin{array}{lll}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

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

$\Delta=\left|\begin{array}{lll}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

$=(5 x+4)\left|\begin{array}{lll}1 & 1 & 1 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$

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

$\Delta=(5 x+4)\left|\begin{array}{llc}1 & 0 & 0 \\ 2 x & -x+4 & 0 \\ 2 x & 0 & -x+4\end{array}\right|$

$=(5 x+4)(4-x)(4-x)\left|\begin{array}{lll}1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1\end{array}\right|$

Expanding along $C_{3}$, we have:

$\begin{aligned} \Delta &=(5 x+4)(4-x)^{2}\left|\begin{array}{ll}1 & 0 \\ 2 x & 1\end{array}\right| \\ &=(5 x+4)(4-x)^{2} \end{aligned}$

Hence, the given result is proved.

(ii) $\Delta=\left|\begin{array}{lll}y+k & y & y \\ y & y+k & y \\ y & y & y+k\end{array}\right|$

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

$\Delta=\left|\begin{array}{lll}3 y+k & 3 y+k & 3 y+k \\ y & y+k & y \\ y & y & y+k\end{array}\right|$

$=(3 y+k)\left|\begin{array}{lll}1 & 1 & 1 \\ y & y+k & y \\ y & y & y+k\end{array}\right|$

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

$\Delta=(3 y+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k\end{array}\right|$

$=k^{2}(3 y+k)\left|\begin{array}{lll}1 & 0 & 0 \\ y & 1 & 0 \\ y & 0 & 1\end{array}\right|$

Expanding along $C_{3}$, we have:

$\Delta=k^{2}(3 y+k)\left|\begin{array}{ll}1 & 0 \\ y & 1\end{array}\right|=k^{2}(3 y+k)$

Hence, the given result is proved.