Question

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

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

Answer 1

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

Given $\alpha ,\beta$ and $\gamma$ are the roots of $x^3-3x^2+3x+7=0$

$x=-1$ is the root of equation $x^3-3x^2+3x+7=0$

$x+1$ should divide $x^3-3x^2+3x+7=0$

$x+1)x^3-3x^2+3x+7(x^2-4x+7$

$x^3+x^2$

_______________

$-4x^2+3x+7$

$-4x^2-4x$

______________

$7x+7$

$7x+7$

___________

x

___________

$(x^3-3x^2+3x+7)=0$

$\Rightarrow (x+1)(x^2-4x+7)=0$

$\Rightarrow x+1=0;x^2-4x+7=0$

$x=-1$ $x=\frac{4\pm \sqrt{16-28}}{2}$

$x=\frac{4\pm \sqrt{12}i}{2}$

$x=2\pm \sqrt{3}i$

Let $\alpha =-1,\beta =2+\sqrt{3}i$, $r=2-\sqrt{3}i$

$\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}$

$=\frac{-2}{1+\sqrt{3}i}+\frac{1+\sqrt{3}i}{1-\sqrt{3}i}-\frac{1-\sqrt{3}i}{+2}$

$=\frac{(-2)(1-\sqrt{3}i)}{4}+\frac{(1+\sqrt{3}ii{)}^2}{4}+\frac{(1+\sqrt{3}i)}{(+2)}$ Multiplying & dividing first term by $(1-\sqrt{3}i)$ & second term by $(1+\sqrt{3}i)$

$=\frac{-2+2\sqrt{3}i+1-3+2\sqrt{3}i-2+2\sqrt{3}i}{4}$

$=\frac{-6+6\sqrt{3}i}{4}$

$=\frac{-3}{2}+\frac{3}{2}\sqrt{3}i=3\times (\frac{-1}{2}+\frac{\sqrt{3}}{2}I)=3{\omega }^2$

$\therefore \frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\alpha -1}+\frac{\gamma -1}{\alpha -1}=3{\omega }^2$.