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?

$$\alpha\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
=\pmatrix{3&-1&-1&0&0\\0&3&-1&-1&0\\0&0&3&-1&-1\\-5&0&0&3&-1\\-5&-5&0&0&3}
\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
$$
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 $\Bbb Q(\alpha)=\Bbb Q(5^{1/5})$.
