If $\alpha\notin\mathbb{Q}$ and $p,q\in\mathbb{Q}$ such that $\alpha^3-15\alpha=q$ and $\alpha^4-56\alpha=p$ then:
(1) Polynomial division of $\alpha^4-56\alpha-p$ by $\alpha^3-15\alpha-q$ gives remainder $15\alpha^2+(p-56)\alpha-q=0$
(2) Polynomial division of $3375(\alpha^4-56\alpha-p)$ by $15\alpha^2+(p-56)\alpha-q$ gives remainder $(p^2-112p+15q-239)q-(p^3-168p^2+30pq+9408p-1680q+13384)\alpha=0$
(3) Polynomial division of $225(\alpha^3-15\alpha-p)$ by $15\alpha^2+(p-56)\alpha-q$ gives remainder $(p^2-112p+15q-239)\alpha-pq-225p+56q=0$
The equation in (2) gives only $p=-4$ and $q=-15$. These are easily verified to obey the equation in (3).
This makes the equation in (1) be $15(\alpha^2-4\alpha+1)=0$ so that $\alpha=2\pm\sqrt{3} $.