Question

If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3-3x^2+3x+7=0$ ($\omega$ is cube root of unity), then $\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}$ is
$(Note:\omega =\frac{-1+i\sqrt{3}}{2})$

• A. $3\omega$
• B. ${\omega }^2$
• C. $2{\omega }^2$
• D. $3{\omega }^2$

Answer 1

The correct option is D $3{\omega }^2$

Let $w$ be the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$, where $w=\frac{-1+i\sqrt{3}}{2}$ $w^2=\frac{-1-i\sqrt{3}}{2}$
$x^3-3x^2+3x+7=0$
${(x-1)}^3+8=0$
$\Rightarrow x=1+2{(-1)}^{\frac{1}{3}}=1+2{(\cos \pi +i\sin \pi )}^{\frac{1}{3}}$
$\Rightarrow x=1+2(\cos \left(\frac{2k\pi +\pi }{3}\right)+i\sin \left(\frac{2k\pi +\pi }{3}\right))$ ...{ De Moivre's Theorem }
where, $k=0,1,2$.
For $k=0$,
$\Rightarrow \alpha =1+2(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right))=1+2\left(\frac{1+i\sqrt{3}}{2}\right)=1-2w^2$
For $k=1$,
$\Rightarrow \beta =1+2(\cos \pi +i\sin \pi )=1-2=-1$
For $k=2$,
$\Rightarrow \gamma =1+2(\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right))=1+2\left(\frac{1-i\sqrt{3}}{2}\right)=1-2w$
$z=\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=\frac{-2w^2}{-2}+\frac{-2}{-2w}+\frac{-2w}{-2w^2}$
$\Rightarrow z=3w^2$
Hence, option D is correct.