Find $\lfloor {\alpha}^6 \rfloor$
Solution 1:
You might start by multiplying the equation by $x$ to get $$\alpha^6=\alpha^4-\alpha^2+2\alpha$$ Next, dividing by $x$ gives $$\alpha^4=\alpha^2-1+\frac2\alpha$$ and combining these two facts we have $$ \alpha^6 = 2\alpha - 1 + \frac 2\alpha $$ whose range on $(1,2)$ is small enough to be dealt with by the floor.