Question

If $\frac{x+1}{x}=3$, then find the value of $\frac{x^6+1}{x^6}=$?

• A. $927$
• B. $114$
• C. $364$
• D. $322$

Answer 1

The correct option is D

$322$

Step 1: Squaring the equation $\frac{x+1}{x}=3$.

Given, $x+\frac{1}{x}=3$

On squaring both sides,

${(\mathrm{x}+\frac{1}{\mathrm{x}})}^2=3^2$

By using identity, ${(a+b)}^2=a^2+b^2+2ab$, we get,

$$\begin{gathered} \Rightarrow {\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)+2\left(\mathrm{x}\right)\left(\frac{1}{\mathrm{x}}\right)=9 \\ \Rightarrow {\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)+2=9 \\ \Rightarrow {\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)=9-2 \\ \Rightarrow {\mathrm{x}}^2+\left(\frac{1}{{\mathrm{x}}^2}\right)=7 \end{gathered}$$

Step 2 : Find the value of the given expression.

Now on cubing both sides, we get,

${[{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}]}^3=7^3$

Using the property, ${(\mathrm{a}+\mathrm{b})}^3={\mathrm{a}}^3+{\mathrm{b}}^3+3\mathrm{ab}(\mathrm{a}+\mathrm{b})$, we get,

$$\begin{gathered} {\Rightarrow \left[{\mathrm{x}}^2\right]}^3+{\left[\frac{1}{{\mathrm{x}}^2}\right]}^3+3\left({\mathrm{x}}^2\right)\left(\frac{1}{{\mathrm{x}}^2}\right)[{\mathrm{x}}^2+(\frac{1}{{\mathrm{x}}^2}\left)\right]=343 \\ \Rightarrow {\mathrm{x}}^6+\left(\frac{1}{{\mathrm{x}}^6}\right)+3\times 7=343\left[\because x^2+\frac{1}{x^2}=7\right] \\ \Rightarrow {\mathrm{x}}^6+\left(\frac{1}{{\mathrm{x}}^6}\right)+21=343 \\ \Rightarrow {\mathrm{x}}^6+\left(\frac{1}{{\mathrm{x}}^6}\right)=343–21 \\ \Rightarrow {\mathrm{x}}^6+\left(\frac{1}{{\mathrm{x}}^6}\right)=322 \end{gathered}$$

Therefore, $\frac{x^6+1}{x^6}=322$

Hence, option (D) is the correct answer.