If α satisfies the equation x ^ 2 + x + 1 = 0 and ( 1 + α ) ^ 7 = A...

We know that, $1 + \omega + {\omega}^{2} = 0$.

So, the roots of the equation ${x}^{2} + x + 1 = 0$ are $\omega$ and ${\omega}^{2}$.

Let, $\alpha = \omega$

$⇒ {\left(1 + \alpha\right)}^{7} = {\left(1 + \omega\right)}^{7}$

$⇒ {\left(1 + \alpha\right)}^{7} = {\left(- {\omega}^{2}\right)}^{7}$

$⇒ {\left(1 + \alpha\right)}^{7} = - {\omega}^{14}$

We know that, ${\omega}^{3n} = 1$.

$⇒ {\left(1 + \alpha\right)}^{7} = - {\omega}^{2} = 1 + \omega$

Now, on comparing with equation we get,

$⇒ A = 1, B = 1, C = 0$

$⇒ 5\left(3A - 2B - C\right) = 5\left(3 - 2 - 0\right)$

$⇒ 5\left(3A - 2B - C\right) = 5$