Question

If $\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3-7x+7=0$, then $\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=$

• A. $\frac{7}{3}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{7}{4}$

Answer 1

The correct option is B

$\frac{3}{7}$

Explanation for the correct option:

Step 1. Find the value of $\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}$:

$\Rightarrow$$\sum \alpha =0,\sum \alpha \beta =–7$ and $\alpha \beta ɣ=–7$

$\Rightarrow$$\sum \alpha \beta =\alpha \beta ɣ$

Also, $\sum \left(\frac{1}{\alpha }\right)=1$

Step 2. By squaring both sides, we get

$\sum {\left(\frac{1}{\alpha }\right)}^2=1$

$\Rightarrow$$\sum \left(\frac{1}{{\alpha }^2}\right)+2\left(\frac{\sum \alpha }{\alpha \beta \gamma }\right)=1$

$\Rightarrow$ $\sum \left(\frac{1}{{\alpha }^2}\right)=1$ $\left[\because \sum \alpha =0\right]$

Step 3. By squaring again both sides, we get

${\left[\sum \left(\frac{1}{{\alpha }^2}\right)\right]}^2=1$

$\Rightarrow$$\sum \left(\frac{1}{{\alpha }^4}\right)+2\left[\frac{\sum {\alpha }^2}{{\left(\alpha \beta \gamma \right)}^2}\right]=1$

$\sum \left(\frac{1}{{\alpha }^4}\right)=1-2\left[\frac{\sum {\alpha }^2}{{\left(\alpha \beta \gamma \right)}^2}\right]$

$$\begin{gathered} =1-2\left[\frac{14}{{\left(-7\right)}^2}\right] \\ =1-\frac{28}{49} \\ =1-\frac{4}{7} \\ =\frac{\mathbf{3}}{\mathbf{7}} \end{gathered}$$

Hence, Option ‘B’ is Correct.