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 $\lfloor a^6 \rfloor = 3$?

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.
