Absolutely Radical

$\sqrt{x^6} = \left| x^3 \right|$ which does NOT always equal $x^3$ since $x^3$ can take negative values.

$\sqrt{x^2} = \left| x \right|$ which does NOT always equal $x$ since $x$ can take negative values.

$\sqrt{x^4} = \left| x^2 \right| = x^2$ because $x^2$ always takes positive values.

$x^{n} = \sqrt{x^{2n} } \text{ must be true for all }x \in \mathbb R \\ x \rightarrow -k ,\quad k\in \mathbb R^{+} \implies (-k)^{n}=\sqrt{(-k)^{2n}} \\ \text{Since RHS is +ve , " n " must be even } \\ \text{The only correct option is } x^2=\sqrt{x^4}$