Question

If $\alpha ,\beta ,\gamma$ are roots of $x^3-3x^2+3x+26=0$ and $\omega$ is cube roots of unity then the value of
$\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}$ equals

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

Answer 1

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

Let $w$ be the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$ where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
$x^3-3x^2+3x+26=0$
${(x-1)}^3+27=0$
$\Rightarrow x=1+3{(-1)}^{\frac{1}{3}}=1+2{(\cos \Pi +i\sin \Pi )}^{\frac{1}{3}}$
$\Rightarrow x=1+3(\cos \left(\frac{2k\Pi +\Pi }{3}\right)+i\sin \left(\frac{2k\Pi +\Pi }{3}\right))$ ...{ De Moivre's Theorem }
where, k=0,1,2.
for k=0
$\Rightarrow \alpha =1+3(\cos \left(\frac{\Pi }{3}\right)+i\sin \left(\frac{\Pi }{3}\right))=1+3\left(\frac{1+i\sqrt{3}}{2}\right)=1-3w^2$
for k=1
$\Rightarrow \beta =1+3(\cos \Pi +i\sin \Pi )=1-3=-2$
for k=2
$\Rightarrow \gamma =1+3(\cos \left(\frac{5\Pi }{3}\right)+i\sin \left(\frac{5\Pi }{3}\right))=1+3\left(\frac{1-i\sqrt{3}}{2}\right)=1-3w$
$z=\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=\frac{-3w^2}{-3}+\frac{-3}{-3w}+\frac{-3w}{-3w^2}$
$\Rightarrow z=3w^2$