assume cube root of 2 is rational.
Then ³√2 = m/n where gcd(m, n) = 1, so 2 = m³/n³ and m³=2n³. Thus, m³ is even so m is even.
Since m is even, let m = 2k. Then m³ = 8k³. So 8k³ = 2n³. So 4k³ = n³ and n³ must be even, so n is even.
But then both m and n are even, so the gcd(m, n) is not 1. Contradiciton.
For the second one, let's do an informal proof. If n² + n + 41 is always prime for any natural number n, then we have a way to calculuate values of primes that are infinitely large. We don't have such a way, therefore the statement is not true.
Now if only math teachers would accept that argument
41*41 + 41 + 41 = 41(41 + 1 + 1). So 41 | n² + n + 41 when n=41