We have, x4−x2+x−1=0 . . . (i) ‌ Let ‌‌‌α=‌
1+√3i
2
,β=‌
1−√3i
2
α+β=‌
1+√3i
2
+‌
1−√3i
2
=1 αβ=(‌
1+√3i
2
)(‌
1−√3i
2
)=‌
1+3
4
=1 Then equation is x2−(α+β)x+αβ=0 ⇒‌‌x2−x+1=0 Now, divide x4−x2+x−1 by x2−x+1, We get, x2−1 ∴ The real roots are x2−1=0 x=±1