The roots of the equation x^3-3x^2+3x+7=0 are | vedclass

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

Question

(D) Given equation: $x^3-3x^2+3x+7=0$. This can be written as $(x-1)^3 + 8 = 0$,so $(x-1)^3 = -8$. Thus,$x-1 = -2, -2\omega, -2\omega^2$. The roots ar

Vedclass · Accepted Answer

$3\omega^2$

Vedclass · Answer

(D) Given equation: $x^3-3x^2+3x+7=0$. This can be written as $(x-1)^3 + 8 = 0$,so $(x-1)^3 = -8$. Thus,$x-1 = -2, -2\omega, -2\omega^2$. The roots are $\alpha = -1, \beta = 1-2\omega, \gamma = 1-2\omega^2$. Let $y = x-h$,so $x = y+h$. Substituting into the equation: $(y+h-1)^3 + 8 = 0$. For the $y^2$ and $y$ terms to be missing,we must have $h-1 = 0$,so $h=1$. The new roots are $\alpha-h = -2, \beta-h = -2\omega, \gamma-h = -2\omega^2$. We need to calculate $S = \frac{\alpha-h}{\beta-h} + \frac{\beta-h}{\gamma-h} + \frac{\gamma-h}{\alpha-h}$. $S = \frac{-2}{-2\omega} + \frac{-2\omega}{-2\omega^2} + \frac{-2\omega^2}{-2} = \frac{1}{\omega} + \frac{1}{\omega} + \omega^2 = \omega^2 + \omega^2 + \omega^2 = 3\omega^2$.

Similar Questions

$\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} \right) - i \cos \left( \frac{2n\pi}{21} \right) \right] = $

If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \operatorname{cis} \theta$,then one of the values of $\sqrt{r \operatorname{cis} \theta}$ is

If $\omega$ is a complex cube root of unity and $a, b, c$ are distinct real numbers,then $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} = $

Express the following expression in the form $A + iB$: $(\cos 2\theta + i\sin 2\theta )^{ - 5} (\cos 3\theta - i\sin 3\theta )^6 (\sin \theta - i\cos \theta )^3$.

What is the product of all the values of $(1-i)^{\frac{2}{5}}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module