Let α,β and γ be the roots of the equation x3−x2+3x+1=0 . Then the value of γ β α+2β+2γα γ β+2α+2γβ α γ+2α+2β is equal to
-10
-8
7
9
Δ=γβα+2β+2γαγβ+2α+2γβαγ+2α+2β Applying C3→C3−2C1−2C2, we get Δ=γβααγββαγ=(α+β+γ)α2+β2+γ2−αβ−βγ−γα=(α+β+γ)(α+β+γ)2−3(αβ+βγ+γα)=1×12−3×(3)=−8