Question

Let $x+\frac{1}{x}=1$ and $a, b$ and $c$ are distinct positive integers such that $\left(x^{a}+\frac{1}{x^{a}}\right)+\left(x^{b}+\frac{1}{x^{b}}\right)+\left(x^{c}+\frac{1}{x^{c}}\right)=0$. Then the minimum value of $(a+b+c)$ is

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

Answer: (C)

$x+\frac{1}{x}=1$
or $x^{2}-x+1=0$
$\therefore x=\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$
or $x=e^{\frac{i \pi}{3}}$
$\therefore x^{a}+x^{-a}$
$=e^{\frac{i a \pi}{3}}+e^{\frac{-i a}{3}}$
$=2 \cos \frac{a \pi}{3}$
Hence, $\cos \frac{a \pi}{3}+\cos \frac{b \pi}{3}+\cos \frac{c \pi}{3}=0$
$a, b, c \in I$
$\therefore a+b+\left.c\right|_{\min }=(1+3+5)=9$