Any irreducible polynomial having $\rm\:k\:$ as a root necessarily has minimal degree. This proof is easy: the set $\rm\:S\:$ of polynomials $\rm\:f \in \mathbb Q[x]\:$ such that $\rm\:f(k) = 0\:$ is closed under addition, and under multiplication by any $\rm\:g\in \mathbb Q[x],\:$ hence it is closed under gcd, since, by Bezout, the gcd of $\rm\:f,g\:$ is a linear combination $\rm\:a\:f + b\:g,\:$ for some $\rm\: a,b\in \mathbb Q[x].\:$ Suppose irreducible $\rm\:f \in S.\:$ Now $\rm\:g\in S\:$ $\Rightarrow$ $\rm\:gcd(f,g)\in S.\:$ Since $\rm\:f\:$ is irreducible, $\rm\:gcd(f,g) = 1\:$ or $\rm\:f.\:$ But $\rm\:1\not\in S\:$ since $\rm\:k\:$ is not a root of $\:1,\:$ so $\rm\:gcd(f,g) = f,\:$ i.e. $\rm\:f\ |\ g.\:$ Since $\rm\:f\:$ divides every $\rm\:g\in S,\:$ we infer $\rm\:f\:$ has minimal degree in $\rm\:S.$ The structure at the heart of this proof will become clearer once you learn some ideal theory.
In your case $\rm\:f(x) = (x^3-7)^2-2 = x^6 - 14\:x^3 + 47\:$ is irreducible over $\rm\mathbb Q$ since it is irreducible mod $5$. Hence it is the minimal polynomial of $\rm\:k\:$ over $\mathbb Q$.