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