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if α and β are the rots of x2+x+1=0 then for y0 in R, y+1αβαy+Β1β1y+α=

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a
yy2−1
b
yy2−3
c
y3−1
d
y3

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detailed solution

Correct option is D

Since α,β are roots of x2+x+1=0, we have α+β=-1 and αβ=1 Now y+1αβαy+β1β1y+α             R1→R1+R2+R3             =y+1+α+Β111αy+Β1Β1y+α   taking y+1+α+β common from R1     =yy2+(α+B)y+αB-1-αy+α2-B+1α-By-B2=yy2-y+1-1-αy-α2+B+α-By-B2∣=yy2-y-y(α+B)+(α+B)-α2+B2=yy2=y3


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