Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Option 1: 4
Option 2: 5
Option 3: 6
Option 4: 8
Correct Answer: 5
Solution : Let $( x+\frac{1}{x})$ be $y$. $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$ Taking $x$ common in both numerator and denominator, we get, $⇒\frac{x( x+\frac{1}{x}-1)}{x( x+\frac{1}{x}+1)}=\frac{2}{3}$ $⇒\frac{y-1}{y+1}=\frac{2}{3}$ $⇒3y-3=2y+2$ $\therefore y = x+\frac{1}{x} = 5$ Hence, the correct answer is 5.
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