I have a 4-torsion point on elliptic curve
$$E : y^2=x^3-3267x+45630$$ given by $(x,y)=(15-36\sqrt{-2},27\sqrt{256\sqrt{-2} - 160})$ which I have checked that it satisfies my elliptic curve. Let $\beta=\sqrt{-2}$. So we have $$(x,y)=(15-36\beta,27\sqrt{256\beta - 160})$$
But than, I was told that the coordinate can be written as $$(x,y)=(15-36\beta,27\alpha(\alpha^2-4\beta-5))$$ where $\alpha^4 - 5\alpha^2 - 32=0$ and $\beta=\sqrt{-2}$.
My question is how is the y-coordinate $27\sqrt{256\beta - 160}$ equals to $27\alpha(\alpha^2-4\beta-5)$ ? How can I obtain that?