lf α and β are the roots of 6x2−6x+1=0, then 12(a+bα+cα2+dα3)+12(a+bβ+cβ2+dβ3)=
If α and β are the roots of 6x2−6x+1, then
α+β=−−66=1
αβ=16
α2+β2=(α+β)2−2αβ=1−13=23
α3+β3=(α+β)3−3αβ(α+β)=1−3×16=12
Now, 12(a+bα+cα2+dα3)+12(a+bβ+cβ2+dβ3)
=12(a+a)+b2(α+β)+c2(α2+β2)+d2(α3+β3)
=a+b2×1+c2×23+d2×12
=a+b2+c3+d4