Question 363976
The cube root of {{{x^6}}} is equal to {{{x^2}}} for any value of x because the sign of x is retained by the presence of the power 2. But {{{sqrt(x^6)}}} can take on only non-negative values (because it is a square root!).  Technically speaking, {{{sqrt(x^6)}}} = |{{{x^3}}}|, and so there would be a problem if x<0.  For example, if we allow x <0, then {{{sqrt((-2)^6) = (-2)^3 = -8}}}, but a square root is always non-negative!