We know that, 1+ω+ω2=0.
So, the roots of the equation x2+x+1=0 are ω and ω2.
Let, α=ω
⇒(1+α)7=(1+ω)7
⇒(1+α)7=(−ω2)7
⇒(1+α)7=−ω14
We know that, ω3n=1.
⇒(1+α)7=−ω2=1+ω
Now, on comparing with equation we get,
⇒A=1,B=1,C=0
⇒5(3A−2B−C)=5(3−2−0)
⇒5(3A−2B−C)=5