(i) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right| = (5x + 4){(4 - x)^2}$
(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right| = {k^2}(3y + k)$

(i) L.H.S. $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + 4}&{2x}&{2x}\\{2x}&{x + 4}&{2x}\\{2x}&{2x}&{x + 4}\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2} + {C_3},$

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

Applying ${R_1} \to {R_1} - {R_2},$

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

Expanding along ${R_1},$

we get
$(5x + 4)\left[ { - (x - 4)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{2x}\\1&{x + 4}\end{array}} \right|} \right]$

$= \left( {5x + 4} \right)\left[ { - \left( {x - 4} \right)\left( {x + 4 - 2x} \right)} \right]$

$= \left( {5x + 4} \right)\left[ { - \left( {x - 4} \right)\left( { - x + 4} \right)} \right] = \left( {5x + 4} \right)\left( {4 - x} \right)\left( {4 - x} \right)$

$= \left( {5x + 4} \right){\left( {4 - x} \right)^2} = R.H.S.$
Hence, proved.

(ii) L.H.S.$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{y + k}&y&y\\y&{y + k}&y\\y&y&{y + k}\end{array}} \right|$

Applying ${C_1} \to {C_1} + {C_2} + {C_3},$

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

Applying ${R_1} \to {R_1} - {R_2},$

we get
$(3y + k)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - k}&0\\1&{y + k}&y\\1&y&{y + k}\end{array}} \right| = (3y + k)\left[ {k\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&y\\1&{y + k}\end{array}} \right|} \right]$

$= (3y + k)k[y + k - y] = (3y + k){k^2} =$ R.H.S.
Hence, proved.