I have difficulty with that. But I can show that squared algebraic number is algebraic.
Let $a$ be algebraic number. Than there is a polynomial $P(x)$ of power $n$ with integer coefficients (the leading coefficient is $1$), and $P(a) = 0$.
Factorizing the $P$ we have: $P(x) = \Pi (x - a_{i})$.
Define polynomial $Q$: $Q(x) = \Pi (x + a_{i})$.
$Q(x)$ has integer coefficients: $Q(x) = \Pi (-((-x) - a_{i})) = (-1)^{deg(P)} P(-x)$.
$P(x) Q(x) = \Pi (x^2 - {a_i}^2)$. So all powers for $x$ in the polynomial $P (x) Q (x)$ are even: $P(x) Q(x) = R(x^2)$. And $R(x)$ has integer coefficients. And $R(a^2) = 0$. So $a^2$ is algebraic.
The same trick works:
Let $Q(x) = \prod (x - a_{i}\omega)(x - a_{i}\omega^2)$, where $\omega$ is a primitive cubic root of unit.
Then $P(x) Q(x) = \prod (x^3 - {a_i}^3)=R(x^3)$ where $R(x)$ has integer coefficients. Then $R(a^3)=0$.