Let α, β be the roots of the equation x ^ 2 - x + 2 = 0 with Im ( α...

Given equation ${x}^{2} - x + 2$ has roots $\alpha \text{ and } \beta$,

So, $\alpha + \beta = 1, \alpha\beta = 2$

$⇒ {\alpha}^{4} + {\beta}^{4} = {\left({\alpha}^{2} + {\beta}^{2}\right)}^{2} - 2{\alpha}^{2}{\beta}^{2}$

$⇒ {\alpha}^{4} + {\beta}^{4} = {\left[{\left(\alpha + \beta\right)}^{2} - 2\alpha\beta\right]}^{2} - 2{\alpha}^{2}{\beta}^{2}$

$⇒ {\alpha}^{4} + {\beta}^{4} = {\left(1 - 4\right)}^{2} - 2 \times 4$

$⇒ {\alpha}^{4} + {\beta}^{4} = 1$

It is given that, $\alpha$ is a root of ${x}^{2} - x + 2$.

$⇒ {\alpha}^{2} - \alpha + 2 = 0$

$⇒ {\alpha}^{2} = \alpha - 2$

$⇒ {\alpha}^{4} = {\alpha}^{2} + 4 - 4\alpha$

$⇒ {\alpha}^{4} = \alpha - 2 + 4 - 4\alpha$

$⇒ {\alpha}^{4} = 2 - 3\alpha$

Now, ${\alpha}^{6} - 5{\alpha}^{2} = {\alpha}^{2}\left({\alpha}^{4} - 5\right)$

$⇒ {\alpha}^{6} - 5{\alpha}^{2} = \left(\alpha - 2\right)\left(2 - 3\alpha - 5\right)$

$⇒ {\alpha}^{6} - 5{\alpha}^{2} = \left(\alpha - 2\right)\left(- 3\alpha - 3\right)$

$⇒ {\alpha}^{6} - 5{\alpha}^{2} = - 3\left({\alpha}^{2} - \alpha - 2\right)$

$⇒ {\alpha}^{6} - 5{\alpha}^{2} = - 3\left(\alpha - 2 - \alpha - 2\right)$

$⇒ {\alpha}^{6} - 5{\alpha}^{2} = 12$

$⇒ {\alpha}^{6} + {\alpha}^{4} + {\beta}^{4} - 5{\alpha}^{2} = 13$