Question

$\alpha$, $\beta$, $\gamma$ are the roots of $x^3-3x^2$ + 3x + 7 = 0(w is cube root of unity) then $(\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1})$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We have $x^3-3x^2$ + 3x + 7 = 0

$\Rightarrow$ $(x-1{)}^3$ + 8 = 0

$\Rightarrow$ ${\left(\frac{(x-1)}{-2}\right))}^3$ = 1

$\Rightarrow$ $\left(\frac{x-1}{-2}\right)$ = 1, $\omega$, ${\omega }^2$

x = -1; 1 - 2 $\omega$; 1 -2${\omega }^2$

$\alpha$ = -1, $\beta$ = 1 - 2$\omega$; $\gamma$ = 1 - 2${\omega }^2$

required expression = 3${\omega }^2$.