Let $\omega=\operatorname{cis}\left(\frac{2 \pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)+i \sin…
Let $\omega=\operatorname{cis}\left(\frac{2 \pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)$ and $f(x)=x^7-2 x^4-4 x^3+8$. Which of the following option is correct?
$\left\{2^{\frac{1}{2}}, 2^{\frac{1}{3}}, \omega, 2^{\frac{1}{3}} \omega\right\}$ is a subset of the solution set of $f(x)$.
$\left\{2^{\frac{1}{2}},-2^{\frac{1}{3}}, 2^{\frac{1}{3}} \omega^2, 2^{\frac{1}{2}} i\right\}$ is a subset of the solution set of $f(x)$.
$\left\{2^{\frac{1}{3}}, 2^{\frac{1}{2}}-2^{\frac{1}{2}}, i, 2^{\frac{1}{3}} \omega^2\right\}$ is not a subset of the solution set of $f(x)$.
$\left\{2^{\frac{1}{3}}, 2^{\frac{1}{3}}, \omega, 2^{\frac{1}{2}} i,-2^{\frac{1}{2}}\right\}$ is a subset of the solution set of $f(x)$.
Solution
Given, $\omega=\operatorname{cis}\left(\frac{2 \pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)$
$f(x)=x^7-2 x^4-4 x^3+8$
$\omega=\frac{-1}{2}+i \frac{\sqrt{3}}{2}=\frac{-1+i \sqrt{3}}{2}$
$f(x)=x^7-2 x^4-4 x^3+8$
$\Rightarrow \quad x^4\left(x^3-2\right)-4\left(x^3-2\right)$
$f(x)=\left(x^3-2\right)\left(x^4-4\right)$
For finding solution, let $f(x)=0$
$\left(x^3-2\right)\left(x^4-4\right)=0$
$\Rightarrow \quad x^3-2=0$ ...(i)
or $\quad x^4-4=0$
$x=2^{1 / 3}$
or $x^4=4$
$x^2= \pm 2$
When $x^2=2$ and when $x^2=-2$
$x= \pm 2^{1 / 2}$ and $x=2^{1 / 2} i$
From Eq. (i), we get
$\begin{array}{r}x^3-2=0 \\ x^3-(\sqrt[3]{2})^3=0\end{array}$
$\Rightarrow \quad\left(x-2^{1 / 3}\right)\left(x^2+2^{2 / 3}+2^{1 / 3} x\right)=0$
$\Rightarrow \quad x-2^{1 / 3}=0$
or $\quad x^2+2^{1 / 3} x+2^{2 / 3}=0$
$\Rightarrow \quad x=2^{1 / 3}$
or $x=\frac{-2^{2 / 3} \pm \sqrt{2^{2 / 3}-4 \cdot 2^{2 / 3}}}{2}$
$\begin{aligned} & x=\frac{-2^{1 / 3} \pm 2^{1 / 3} \sqrt{3} i}{2} \\ & x=\frac{2^{1 / 3}(-1+\sqrt{3} i)}{2}=2^{1 / 3} \omega\end{aligned}$
$\therefore$ Solution set $=\left\{2^{1 / 3}, 2^{1 / 3} \omega, 2^{1 / 2} i,-2^{1 / 2}\right\}$.