If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$. 
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.

Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?

Let $f(x)=x^5-x^3+x-2$.


*

*First step. Differenciate the polynomial to obtain:
$$f'(x)=5x^4-3x^2+1$$
which has no zeros. Hence, $f$ has exactly one root.

*Second step. We have that $f(1)=-3<0$ and $f(2)=24>0$. Then the root lies in the interval $(1,2)$.

*Third step. Let $a$ be the root of $f$. We have
$$a^6=a\cdot a^5=a(a^3-a+2)$$
Now define $g(x)=x(x^3-x+2)$. Also differentating, we can see easily that $g$ is increasing in $[1,2]$, so now you have to try more values on $f$, to get finer bounds for the root. If we have that $l<a<u$ and $\lfloor g(l)\rfloor=\lfloor g(u)\rfloor$, then $\lfloor a^6 \rfloor$ is also the same.
