Given $x^5-x^3+x-2=0$, find $\lfloor x^6\rfloor$. 
If $\alpha$ is a real root of the equation $x^5-x^3+x-2=0$, find the value of $\lfloor\alpha^6\rfloor$, where $\lfloor x\rfloor$ is the least positive integer not exceeding $x$.

My approach is to bound the value of $\alpha^6=\alpha^4-\alpha^2+2\alpha$.
First I proved the function $f(x)=x^5-x^3+x-2$ is monotone increasing by derivative. Then I argue that $1<\alpha<\frac32$ as $f(1)<0$ and $f(\frac32)>0$.
Then I tried to create an upper and lower bound for $\alpha$, as such
$$\alpha^6=\alpha^4-\alpha^2+2\alpha<\frac94\alpha^2-\alpha^2+2\alpha=\frac54\alpha^2+2\alpha<\frac{45}{16}+3=\frac{93}{16}$$
and
$$\alpha^6=\alpha^4-\alpha^2+2\alpha>\alpha^3-\alpha^2+2\alpha=\alpha^2(\alpha-1)+2\alpha>2$$
Now we know that $\lfloor\alpha\rfloor\in\{2,3,4,5\}$. But I cannot proceed any further.
Any idea, help, hint, or answer is appreciated. Thank you.

$$x^5-x^3+x-2=x^5+x^2-(x^2-x+1)-(x^3+1)=$$
$$=(x^2-x+1)(x^3+x^2-x-2).$$
Now, we see that our equation has an unique real root $1<\alpha<2$ and from here $$[\alpha]=1.$$
Now, about your new problem.
Easy to see that for our root $\alpha$ we have $$1.205<\alpha<1.206,$$
which gives $$3<\alpha^6<4$$ and from here: $$[\alpha^6]=3.$$
