Given, $\mathrm{\Delta }=\left|\begin{array}{ccc}{\mathrm{a}}^2(1+\mathrm{x})& \mathrm{ab}& \mathrm{ac}\\ \mathrm{ab}& {\mathrm{b}}^2(1+\mathrm{x})& \mathrm{be}\\ \mathrm{ac}& \mathrm{be}& {\mathrm{c}}^2(1+\mathrm{x})\end{array}\right|$

$\Rightarrow \mathrm{\Delta }=\mathrm{abc}\left|\begin{array}{ccc}\mathrm{a}(1+\mathrm{x})& \mathrm{a}& \mathrm{a}\\ \mathrm{b}& \mathrm{b}(1+\mathrm{x})& \mathrm{b}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}(1+\mathrm{x})\end{array}\right|$

[divide column C₁, C₂ and C₃ with a, b, c respectively]

$\Rightarrow \mathrm{A}={\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2\left|\begin{array}{ccc}(1+\mathrm{x})& 1& 1\\ 1& (1+\mathrm{x})& 1\\ 1& 1& (1+\mathrm{x})\end{array}\right|$

[divide row R₁, R₂ and R₃ with a, b, c respectively]

$\Rightarrow \mathrm{A}={\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2\left|\begin{array}{ccc}\mathrm{x}& 0& 1\\ -\mathrm{x}& \mathrm{x}& 1\\ 0& -\mathrm{x}& 1+\mathrm{x}\end{array}\right|\left[\begin{array}{l}{\mathrm{c}}_1\to {\mathrm{c}}_1-{\mathrm{c}}_2\\ {\mathrm{c}}_2\to {\mathrm{c}}_2-{\mathrm{c}}_3\end{array}\right]$

⇒ Δ = a²b²c²[x{(x(1 + x)} + x) – 0 + 1{x² – 0}

⇒ Δ = a²b²c²[x(x + x² + x) + x²]

⇒ Δ = a²b²c²[2x² + x³ + x²]

⇒ Δ = a²b²c²x²(1 + 2 + x)

⇒ Δ = a²b²c²x²(3 + x)

Hence, it is devisable by x².
So, the answer is option C