If α satisfies the equation x 2 + x + 1 = 0 and ( 1 + α ) 7 = A + B α + C α 2 , A , B , C ≥ 0 , then 5 ( 3 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, $\alpha =\omega$

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

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

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

We know that, ${\omega }^{3n}=1$.

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

Now, on comparing with equation we get,

$\Rightarrow \mathrm{A}=1, \mathrm{B}=1,\mathrm{C}=0$

$\Rightarrow 5\left(3\mathrm{A}-2\mathrm{B}-\mathrm{C}\right)=5\left(3-2-0\right)$

$\Rightarrow 5\left(3\mathrm{A}-2\mathrm{B}-\mathrm{C}\right)=5$