Solution
= Solution
Over $\mathbb F_5$,
$$
x^3+3=(x-3)(x^2+3x+4).
$$
The quadratic discriminant is $9-16\equiv3$, while the squares modulo five are $0,1,4$. Thus that factor is irreducible. Adjoining either quadratic root creates a degree-two extension and supplies the other root by the sum-of-roots relation. Consequently the <splitting field> is $\mathbb F_{25}$ and $\boxed{[K:\mathbb F_5]=2}$.