why only square root approach to check number is prime Why do we use only square root approach to find a number is prime or not?
why not cube root & 4rth root?

Consider the number $143$. This number is composite, not prime, because
$$11 \times 13 = 143.$$
But 
$$\sqrt[3]{143} \approx 5.229 < 11 < 13,$$
so if you test only possible factors up to $\sqrt[3]{143}$
then you will not check whether $11$ or $13$ divide $143$,
and you will not discover that $143$ is not prime.
On the other hand, $$11 < \sqrt{143} \approx 11.958,$$
so if you test all possible prime factors less than $\sqrt{143}$
then you will (eventually) test whether $11$ divides $143$,
and you will find out that $143$ is not prime.
Moreover, as other answers have shown, this always works,
even when you replace $143$ by some arbitrary positive integer $n$.
One counterexample is enough to show that an algorithm is
incorrect. Sometimes one can refine an algorithm in the face of such a
counterexample, for example here you might say you will apply the algorithm
only to test primality of a number $n$ greater than $143$.
But there are plenty of other, much larger, counterexamples—just take
$n$ to be the product of any two primes that are "close" to 
each other—and no practical way to account for all the counterexamples 
(as there appear to be infinitely many of them).
More precisely, I do not see any practical way that is easier than 
actually finding out (using a correct algorithm) 
whether the number $n$ is prime.
So the reason we use square root and not cube root is that
the square root approach works and the cube root approach does not work.
The fourth root approach also does not work, since $\sqrt[4]{n} < \sqrt[3]{n}$
for $n > 1$ and we already can see that $\sqrt[3]{n}$ is not always large enough.
