Factorize x^3+3 Task: Factorize $x^3+3$ in $\mathbb{R}$ and $GF(7)$ I think the solution is $x^3+3$ in both cases, so the polynomial already is irreductible. Is my assuption is right, how do i show that?

To check whether a cubic polynomial is irreducible over a given field, it is sufficient to check whether $f$ has any roots in that field (why?).

In $GF(7)$ this is easy: there are only $7$ possible roots, so you can simply evaluate $f(x)$ for all $x \in GF(7)$.

Over $\mathbb{R}$, remember that every polynomial of odd degree has at least one root. You can prove this using intermediate value theorem, since polynomials are continuous functions. Given this, when does $x^3 + 3 = 0$ in $\mathbb{R}$? Once you've found a root, you can use polynomial long division to find a factorization.