If alpha is a non-real root of x^6=1,then fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$. This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$. Since $\alpha$ i

Vedclass · Accepted Answer

$-\alpha^2$

Vedclass · Answer

(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$. This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$. Since $\alpha$ is a non-real root of $x^6=1$,it satisfies the equation $x^5+x^4+x^3+x^2+x+1=0$. Therefore,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$. We can rearrange the terms as: $\alpha^5+\alpha^3+\alpha+1 = -(\alpha^4+\alpha^2)$ $\alpha^5+\alpha^3+\alpha+1 = -\alpha^2(\alpha^2+1)$ Dividing both sides by $(\alpha^2+1)$,we get: $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1} = -\alpha^2$.

If $\alpha$ is a non-real root of $x^6=1$,then $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}$ is equal to

Similar Questions

One of the values of $(\sqrt{3}-i)^{\frac{2}{5}}$ is

If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $P(x) = f(x^3) + xg(x^3)$ is divisible by $x^2 + x + 1$,then $P(1)$ is equal to ....... .

If $\omega$ is a complex cube root of unity,then $(1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \dots$ to $2n$ factors is equal to:

If $z=\frac{\sqrt{3}+i}{2}$,then $\left(z^{101}+i^{103}\right)^{105}=$

$(1+\sqrt{5}+i \sqrt{10-2 \sqrt{5}})^5=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module