Question

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

• A. $-\frac{1}{2}$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

$-1$

Explanation for the correct option:

Step 1. Find the value of $\sum \frac{1}{\alpha +\beta -\gamma }$:

Given, $\alpha ,\beta$ and $\gamma$ are the roots of $x^3-2x+1=0$

$x^3–x–x+1=0$

$\Rightarrow$$x(x^2–1)–(x–1)=0$

$\Rightarrow$ $x=1,x^2+x–1=0$

$\Rightarrow$$\alpha =1,\beta =\frac{-1+\sqrt{5}}{2},\gamma =\frac{-1-\sqrt{5}}{2}$

Step 2. Find the value of $\sum \alpha ,\sum \alpha \beta ,\alpha \beta \gamma$:

$\begin{array}{rcl}\sum \alpha & =& \alpha +\beta +\gamma \\ & =& 1+\frac{-1+\sqrt{5}}{2}+\frac{-1-\sqrt{5}}{2}\\ & =& 0\end{array}$

$\begin{array}{rcl}\sum \alpha \beta & =& \alpha \beta +\beta \gamma +\gamma \alpha \\ & =& 1\times \frac{-1+\sqrt{5}}{2}+\frac{-1+\sqrt{5}}{2}\times \frac{-1-\sqrt{5}}{2}+\frac{-1-\sqrt{5}}{2}\times 1\\ & =& \frac{-8}{2}\\ & =& -4\end{array}$

$\begin{array}{rcl}\alpha \beta \gamma & =& 1\times \frac{-1+\sqrt{5}}{2}\times \frac{-1-\sqrt{5}}{2}\\ & =& \frac{-4}{2}\\ & =& -2\end{array}$

Step 3 . Put the values of $\sum \alpha ,\sum \alpha \beta ,\alpha \beta \gamma$ in $\sum \frac{1}{\alpha +\beta -\gamma }$, we get

$\begin{array}{rcl}\therefore \sum \frac{1}{\alpha +\beta –\gamma }& =& \sum \frac{1}{-\gamma –\gamma }\\ & =& \sum \frac{1}{-2\gamma }\\ & =& -\frac{1}{2}\left(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\right)\\ & =& -\frac{1}{2}\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)\\ & =& -\frac{1}{2}\times \frac{\sum \alpha \beta }{\alpha \beta \gamma }\\ & =& -\frac{1}{2}\times \frac{-4}{-2}\\ & =& \mathbf{-}\mathbf{1}\end{array}$

Hence, Option ‘B’ is Correct.