If α and β are the root of x2+x+1=0, then α16+β16=?
A
2
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B
1
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C
−1
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D
1+√3i2
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Solution
The correct option is D1+√3i2 x2+x+1=0λ16+β16=2Distriminent=b24ac=1−4=−3Roots=−1±√−32⇒λ=(1+√−3)2=(1+√3i)2−w⇒β=√−3−12=(√3i−1)2=wwhere(w)iscuberootsofunity∴1+w+w2=0andw3=1⇒∴λ16+β16=(w2)16=w16(w2+1)⇒w15w(w2+1)=1(w3+w)=1+w=1√3i−12∴1+√3i2Ans.