If $\alpha$  is a real root of the equation  $x^5-x^3+x-2=0$, then find  the value of  $\left[{\alpha }^6\right]$. (For any  real number a,$\left[\text{a}\right]$ represent the greatest integer not exceeding a) 

Suppose a is a real root of the given equation. Then
${\alpha }^5-{\alpha }^3+\alpha -2=0.$    …(1)
This gives  ${\alpha }^5-{\alpha }^3+\alpha -1=1$ and hence   $(\alpha -1)({\alpha }^4+{\alpha }^3+1)$  = 1. Observe that 
 $a^4+a^3+1^{3}2a^2+a^3=a^2(a+2)$
If $-1\le a<0$, then $a+2>0$, giving $a^2(a+2)>0$ and hence $(a-1)(a^4+a^3+1)<0$. 
If $\text{a<\hspace{0.17em}-1}$, then ${\alpha }^4+{\alpha }^3={\alpha }^3(\alpha +1)>0$ and hence  ${\alpha }^4+{\alpha }^3+1>0$.
 This again gives $(a-1)(a^4+a^3+1)<0$.
The above reasoning shows that for $a<0$, we have $a^5-a^3+a-1<0$ and hence cannot be  equal to 1. We conclude that a real root a of $x^5-x^3+x-2=0$ is positive 
(obviously $\alpha \ne 0$ ).
Now using $a^5-a^3+a-2=0,$ we get
 ${\alpha }^6={\alpha }^4-{\alpha }^2+2\alpha$
The statement $\left[{\alpha }^{\text{6}}\right]=\text{ 3}$ is equivalent to  $3\le {\alpha }^6<4$.
Consider ${\alpha }^4-{\alpha }^2+2\alpha <4$ . Since  $\alpha >0$, this is equivalent to  ${\alpha }^5-{\alpha }^3+2{\alpha }^1<4\alpha$.  Using the relation (1), we can write  $2{\alpha }^2-\alpha +2<4\alpha$ or $2{\alpha }^2-5\alpha +2<0$ . Treating this  as a quadratic, we get this is equivalent to $\frac{1}{2}<\alpha <2$ . Now observe that if $\alpha \ge 2$  then  $1=(\alpha -1)({\alpha }^4+{\alpha }^3+1)\ge 25$ which is impossible. If  $0<\alpha \le \frac{1}{2},$ then 1 =   $(\alpha -1)$$({\alpha }^4+{\alpha }^3+1)<0$   which again is impossible. We conclude that $\frac{1}{2}<\alpha <2$ .
Similarly,${\alpha }^4-{\alpha }^2+2\alpha \ge 3$   is equivalent to ${\alpha }^5-{\alpha }^3+2{\alpha }^2-3\alpha \ge 0$  which is equivalent  to  $2{\alpha }^2-4\alpha +2\ge 0$. But this is $2{(\alpha -1)}^2\ge 0$  which is valid. Hence $3\le {\alpha }^6<4$  and we  get $\left[{\alpha }^{\text{6}}\right]=3$.