$\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b} $ just for a,b>0 The property $\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b} $ is valid just if $a,b>0$. If this restriction didn't apply $\sqrt{-1}\sqrt{-1} = -1$ would be equal to $\sqrt{(-1)(-1)} = 1$
Beside giving particular examples, how can I show that the $a,b>0$ is required for $\sqrt[n]{ab} = \sqrt[n]{a}\sqrt[n]{b} $  to be true?
 A: The answer to this question really comes down to the definition of $\sqrt[n]{a}$. If this symbol describes any number $x$ so that $x^n = a$, then $\sqrt[n]a$ describes a set. $\sqrt{-1}\sqrt{-1} = \pm i^2 = \pm 1 = \{-1.1\}$, while $\sqrt{(-1)(-1)} = \sqrt{1} = \pm 1 = \{-1,1\}$. So, in fact, the set of values defined by these two expressions are equivalent. 
You may want to define $\sqrt[n]a$ as the unique positive real number $x, x>0$ when $a>0$ and the unique positive imaginary number $ix, x>0$ when $a<0$. Then, if $a,b<0$, then $ab>0$, so $\sqrt[n]{ab}>0$ by definition. $\sqrt[n]a\sqrt[n]b = xi\cdot yi = -xy <0$ on the other hand, hence the discrepancy. 
When you put a restriction on the definition for $\sqrt[n]a$, you're going to have restrictions on it's properties as well.
A: apply the nth power to both sides and conclude in a true statement then the first statement must be true because of the double implication in an equality. there is no problem because you have the nth root of two positive numbers and the nth root of the product of two positive numbers which makes it also positive. what is more, you can distribute the n in the right hand side of the equation because of the commutative property.
