This will be a proof by contradiction. Assume that 5^(1/3) is rational. Observe that 5^(1/3) is a root of x^3 = 5 which simplifies to x^3 - 5 = 0. Then, by the rational root theorem, 5^(1/3) = p/q where p divides 5 and q divides one. Therefore, the possible values of 5^(1/3) are 5/1, 5/-1, 1/1, 1/-1, -5/1, -5/-1, 1/1, and 1/-1 = 5, -5, 1, -1. However, 5^(1/3) ≠ -1 since (-1)^(3) = -1 ≠ -5, 5^(1/3) ≠ -5 since (-5)^(3) = -125 ≠ 5, 5^(1/3) ≠ 5 since 5^(3) = 125 ≠ 5, and 5^(1/3) ≠ 1 since 1^(3) = 1 ≠ 5. This means that 5^(1/3) can't be a rational root of x^3 - 5 = 0 which does not work. This means that our previous claim that 5^(1/3) is rational does not work and that 5^(1/3) is irrational. Hope this helps! Please tell me if there are any errors in my solution.