Please suggest a method to solve this?
Let $E$ be a field and let $H$ be a subgroup of $Aut(E)$. Let $F$ be the fixed field of $E$ under $H$ and let $f(X) ∈ E[X]$ be a monic polynomial that splits in $E$. Consider the statement P($\alpha$) about an element $\alpha ∈ E$:
P($\alpha$): If f ($\alpha$) = 0, then f (h$\alpha$) = 0 for each h ∈ H.
Prove that if P ($\alpha$) is true for all $\alpha ∈ E$, then $f (X) ∈ F [X]$.
Observe that the coefficients of $\;f\;$ are the symmetric polynomials in its roots (e.g., the free
coefficient is $\;\pm a_0\cdot a_1\cdot...\cdot a_n\;$ , with $\;a_i\;$ the roots of $\;f\;$ ). Since every $\;h\in H\;$ is $\;1-1\;$ and onto
and it maps a roots of $\;f\;$ to a root of $\;f\;$ , all the coefficients of $\;f\;$ are fixed by any
$\;h\in H\iff f(x)\in F[x]=\text{Fix}_H(Aut(E))[x]\;$