Raising The Power of $\alpha$

Let $f(x)=x^5-x^3+x-2$ , which has a root $\alpha$

Given Equation: $x^5-x^3+x-2=0$

We can multiply the equation by $x$ and substitute $x=\alpha$ to get

$\alpha^6=\alpha^4-\alpha^2+2\alpha$
Next, dividing the equation by $x$ and substituting $x=\alpha$ gives

$\alpha^4=\alpha^2-1+\frac2\alpha$
Now, combining the above two facts we get

$\alpha^6 = 2\alpha - 1 + \frac 2\alpha$

Also, $f(1) \times f(2) < 0$

$\implies 1 < \alpha < 2$ {Using Intermediate Value Theorem}

Further, it can be clearly seen by Calculus or otherwise that $f(x)$ is throughout increasing function.

Therefore, $x=\alpha$ is the only real root.

Clearly in the given domain of $\alpha$ { i.e. $1 < \alpha < 2$ } ,

$3 < 2\alpha - 1 + \frac 2\alpha \ < 4$
$\implies 3 < \alpha^6 < 4$
$\implies \lfloor {\alpha}^6 \rfloor = \boxed {3}$

For your information,

(x^5 - x^3 + x - 2) = (x^2 - x + 1)(x^3 + x^2 - x - 2) = 0

alpha = 1.2055694304005903 and

[3.0701059427181745239483510319284] = 3