If alpha is a root of the equation x^2-x+1=0, | vedclass

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

Question

(C) Given the equation $x^2-x+1=0$. The roots are $\alpha = -\omega$ and $\alpha = -\omega^2$,where $\omega$ is the complex cube root of unity. Since

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given the equation $x^2-x+1=0$. The roots are $\alpha = -\omega$ and $\alpha = -\omega^2$,where $\omega$ is the complex cube root of unity. Since $\alpha^2-\alpha+1=0$,we have $\alpha+\frac{1}{\alpha} = 1$. Let $S_n = \alpha^n + \frac{1}{\alpha^n}$. For $n=1$,$S_1 = \alpha + \frac{1}{\alpha} = 1$. For $n=2$,$S_2 = \alpha^2 + \frac{1}{\alpha^2} = (\alpha+\frac{1}{\alpha})^2 - 2 = 1^2 - 2 = -1$. For $n=3$,$S_3 = \alpha^3 + \frac{1}{\alpha^3} = (-1)^3 + \frac{1}{(-1)^3} = -1 - 1 = -2$. For $n=4$,$S_4 = \alpha^4 + \frac{1}{\alpha^4} = \alpha + \frac{1}{\alpha} = 1$. For $n=5$,$S_5 = \alpha^5 + \frac{1}{\alpha^5} = \alpha^2 + \frac{1}{\alpha^2} = -1$. For $n=6$,$S_6 = \alpha^6 + \frac{1}{\alpha^6} = 1 + 1 = 2$. The terms are $S_n^3$. $S_1^3 = 1^3 = 1$,$S_2^3 = (-1)^3 = -1$,$S_3^3 = (-2)^3 = -8$,$S_4^3 = 1^3 = 1$,$S_5^3 = (-1)^3 = -1$,$S_6^3 = 2^3 = 8$. The sequence of terms is $1, -1, -8, 1, -1, 8, 1, -1, -8, 1, -1, 8$. Summing these $12$ terms: $(1-1-8) + (1-1+8) + (1-1-8) + (1-1+8) = -8 + 8 - 8 + 8 = 0$.

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

Similar Questions

For $x \in R$,the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + x \sin \theta - 2 \sin \theta = 0$,where $\theta \in (0, \frac{\pi}{2})$,then the value of $\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to

If $p$ and $q$ are positive real numbers such that $p^{2} + q^{2} = 1$,then the maximum value of $(p+q)$ is:

If $\left|\frac{x^2+kx+1}{x^2+x+1}\right| < 3$ for all real $x$,then $k$ is in the interval

If $\alpha, \beta$ are the irrational roots of the equation $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$,then the roots of the equation $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module