Question 274702
Recall that {{{x^(m/n)^""=root(n,x^m)}}}



This means that {{{(16a^6)^(3/4)^""=root(4,(16a^6)^3)}}}



We could optionally cube {{{16a^6}}} to get {{{4096a^18}}} which would mean that {{{(16a^6)^(3/4)^""=root(4,4096a^18)}}}