Is $\alpha=3-\sqrt[5]{5}-\sqrt[5]{25}$ algebraic over $\mathbb{Q}$? How to determine whether the given real number $\alpha =3-\sqrt[5]{5}-\sqrt[5]{25}$ is algebraic or not. And, $[\mathbb{Q}(\alpha):\mathbb{Q}]=?$
Let $x=3-\sqrt[5]{5}-\sqrt[5]{25}$, 
\begin{align*}
& x = 3-\sqrt[5]{5}-\sqrt[5]{25}\\
\implies & (x-3)  = -\sqrt[5]{5}-\sqrt[5]{25}\\
\implies & (x-3)^5=(-\sqrt[5]{5}-\sqrt[5]{25})^5
\end{align*}
But, solving this is abit tedios process, is there any other way to do this?
 A: $$\alpha\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
=\pmatrix{3&-1&-1&0&0\\0&3&-1&-1&0\\0&0&3&-1&-1\\-5&0&0&3&-1\\-5&-5&0&0&3}
\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
$$
so that $\alpha$ is an eigenvalue of
$$\pmatrix{3&-1&-1&0&0\\0&3&-1&-1&0\\0&0&3&-1&-1\\-5&0&0&3&-1\\-5&-5&0&0&3}.
$$
This matrix has integer entries, and so its characteristic equation
has integer coefficients. Therefore $\alpha$ is an algebraic integer.
As $\alpha\in\Bbb Q(5^{1/5})$ and $\alpha\notin\Bbb Q$ and $|\Bbb Q(5^{1/5}):\Bbb Q|=5$ and $5$ is prime, then $\Bbb Q(\alpha)=\Bbb Q(5^{1/5})$.
A: Since $X^5-5$ is $\mathbb Q$-irreducible (Eisenstein) it is the minimal polynomial of $r:=\sqrt [5] 5$ over $\mathbb Q$.
Hence the field $\mathbb Q(r)=\mathbb Q[r]\cong \mathbb Q[X]/\langle X^5-5\rangle $ is a $\mathbb Q$-vector space with basis $1,r,r^2,r^3,r^4$.
Obviously $\mathbb Q\subset \mathbb Q[\alpha] \subset \mathbb Q[r]$ and since $\alpha =3-r-r^2\notin \mathbb Q$, we have $\mathbb Q\varsubsetneq  \mathbb Q[\alpha] \subset \mathbb Q[r]$.
But the field $\mathbb Q[r]$ has degree $5$ over $\mathbb Q$, a prime number, so that its only non trivial subfield is itself and $\mathbb Q[\alpha] = \mathbb Q[r]$.   Thus  $$[\mathbb Q[\alpha]:\mathbb Q]==[\mathbb Q[r]:\mathbb Q]=5$$
