Question

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3 - 3x^2 + 3x + 7 = 0$, and w is cube root of unity, then the value of $\frac{\alpha-1}{\beta-1} + \frac{\beta-1}{\gamma-1} + \frac{\gamma-1}{\alpha-1}$ is equal to

• A. $3 w^2$
• B. $3/w$
• C. $2w^2$
• D. none of these

Answer 1

The Correct Option is A

We have,
$x^{3}-3x^{2}+3x+7=0$
$\left(x + 1\right)\left(x^{2} - 4x + 7\right) = 0$
$x+1=0$ or $x^{2}-4x+7=0$
$x=-1$ or $x=\frac{4 \pm\sqrt{16-28}}{2}$
$\Rightarrow x=-1$ or $x=\frac{4\pm2\sqrt{3}i}{2}$
$\Rightarrow x=-1$ or $x=2 \pm\sqrt{3}i$
$\Rightarrow x=-1$ or $x=1-2w$, $1-2w, 1-2w^{2}$
$\left[2w=-1+\sqrt{3}i, 2w^{2}=-1-\sqrt{3}i\right]$
$\therefore \alpha=-1, \beta=1-2w$ and $\gamma=1-2w^{2}$
Now, $\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}$
$=\frac{-1-1}{1-2w-1}+\frac{1-2w-1}{1-2w^{2}-1}=\frac{1-2w^{2}-1}{-1-1}$
$=\frac{2}{2w}+\frac{2w}{2w^{2}}+\frac{2w^{2}}{2}$
$=\frac{1}{w}+\frac{1}{w}+w^{2}=\frac{2}{w}+w^{2}$
$=2w^{2}+w^{2}=3w^{2}$