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 $\lfloor a^6 \rfloor = 3$?
 A: We have $\sqrt[6]{3} \approx 1.2009$ and $\sqrt[6]{4} \approx 1.2599$.  Let $f(x)=x^3+x^2-x-2$.  Then $f(1.2) \approx -0.032$ and $f(1.25) \approx 0.2656$.  So, $a$ must be between $1.2$ and $1.25$.
EDIT:  As I said in the comments below, I see no way of showing that there is only one root using only precalculus.  But, for completeness of my answer:  If there were two roots, then the Mean Value Theorem would imply that the derivative is $0$ between the roots.  But, $f'(x) = 3x^2+2x-1$.  The roots of this are $x=-1$ and $x=\frac{2}{3}$.  Using whichever method one likes, you can see that there is a local maximum at $x=-1$ and a local minimum at $x=\frac{2}{3}$.  And, both $f(-1)$ and $f\left(\frac{2}{3}\right)$ are negative.  So, there cannot be another root.
A: 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.
A: Opposed to all of these answers, I will simply go ahead and solve the cubic:
$$0=x^3+x^2-x-2$$
Applying the cubic formula, I get:
$$x=\frac16\left(-2+\sqrt[3]{4}(\sqrt[3]{42+3\sqrt{177}})+\sqrt[3]{42-3\sqrt{177}})\right)$$
According to Wolfram|Alpha.
Then I guess you would either do the floor function after getting it in decimal form or you could somehow take the floor function now.
