$\left| {\begin{array}{cccccccccccccccccccc}{x + 4}&x&x\\x&{x + 4}&x\\x&x&{x + 4}\end{array}} \right|$

We have, $\left| {\begin{array}{cccccccccccccccccccc}{x + 4}&x&x\\x&{x + 4}&x\\x&x&{x + 4}\end{array}} \right| = \left| {\begin{array}{cccccccccccccccccccc}{2x + 4}&{2x + 4}&{2x}\\x&{x + 4}&x\\x&x&{x + 4}\end{array}} \right|$

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

[here, given determinant is expressed in sum of two determinants]

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

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

Applying ${C_1} \to {C_1} - {C_3}$

and ${C_2} \to {C_2} - {C_3}$

in first and applying ${C_1} \to {C_1} - {C_2}$ in second,

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

Expanding both the along first column,

we get
$2x[ - 4( - 4)] + 4[4(x + 4 - 0)]$
$= 2x \times 16 + 16(x + 4)$

$= 32x + 16x + 64$

$= 16(3x + 4)$