118186
If \(\alpha\) is a non-real root of the equation \(x^6-1=0\), then \(\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}\) is equal to
1 \(\alpha\)
2 1
3 0
4 -1
Explanation:
D Given, \(\alpha\) be a non real root of the equation \(x^6-1=0\) Then, \(\begin{array}{ll} \frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}=\frac{\alpha^2(\alpha+1)+\alpha^4(1+\alpha)}{\alpha+1}=\alpha^2+\alpha^4=? \\ \because \quad \alpha \text { is a root of } x^6-1=0 \\ \therefore \quad \alpha^6-1=0 \\ \text { or } \quad \left(\alpha^2\right)^3-1=\left(\alpha^2-1\right)\left(\alpha^2+\alpha^4+1\right)=0 \\ \because \quad \alpha \text { is a non real root. } \\ \therefore \quad \alpha^2-1 \neq 0 \\ \therefore \quad \left(\alpha^2+\alpha^4+1\right)=0 \\ \therefore \quad \alpha^2+\alpha^4=-1\end{array}\)