Yes indeed -- the standard proof of the fact that 2^(1/2) can be mimicked to obtain this result.
Suppose that 5^(1/3) is rational: thus, we may write 5^(1/3) = p/q, where gcd(p, q) = 1. Then 5 = (p^3)/(q^3), or 5q^3 = p^3. Deduce that 5 divides p, that 5 divides q, and hence derive a contradiction.
Suppose that 5^(1/3) is rational: thus, we may write 5^(1/3) = p/q, where gcd(p, q) = 1. Then 5 = (p^3)/(q^3), or 5q^3 = p^3. Deduce that 5 divides p, that 5 divides q, and hence derive a contradiction.