TITLE: Find $\lfloor {\alpha}^6 \rfloor$
QUESTION [7 upvotes]: If $\alpha$ is a real root of the equation
  $$x^5-x^3+x-2=0$$
  find the value of $\lfloor {\alpha}^6 \rfloor$.   

This one totally stumped me. We are asked to calculate $\lfloor {\alpha}^6 \rfloor$ without actually calculating the root or using wolfram alpha or any other calculator. I found that the above equation has only one real root by sketching its graph using calculus. I was also able to use the intermediate value theorem to conclude that $1<\alpha<2$ , but this is of little use while calculating $\lfloor {\alpha}^6 \rfloor$. Please help!

REPLY [20 votes]: 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.