We have,
$x^5-8 x^4+25 x^3-38 x^2+28 x-8=0$
Let
$\begin{aligned}
& f(x)=x^5-8 x^4+25 x^3-38 x^2+28 x-8 \\
& f(2)=32-128+200-152+56-8 \\
& f(2)=0 \Rightarrow f^{\prime}(x)=5 x^4-32 x^3+75 x^2-76 x+28 \\
& f^{\prime}(2)=80-256+300-152+28 \Rightarrow f^{\prime}(2)=0 \\
& f^{\prime \prime}(x)=20 x^3-96 x^2+150 x-76 \\
& f^{\prime \prime}(2)=160-384+300-76 \Rightarrow f^{\prime \prime}(2)=0 \\
& f^{\prime \prime \prime}(x)=60 x^2-192 x+150 \\
& f^{\prime \prime \prime}(2)=240-384+150=6 \Rightarrow f^{\prime \prime \prime}(2) \neq 0
\end{aligned}$
$\therefore$ Given, $\alpha$ is a root of multiplicity 3
$\therefore \quad \alpha=2 \Rightarrow \alpha^2-5 \alpha+6=4-10+6=0$