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 $$\left( {a + b + c} \right)$$ is
A.
7
B.
8
C.
9
D.
10
Answer :
9
Solution :
$$\eqalign{
& x + \frac{1}{x} = 1\,\,\,\,{\text{or }}{x^2} - x + 1 = 0 \cr
& \therefore x = \frac{1}{2} \pm i\frac{{\sqrt 3 }}{2}\,\,\,\,\,{\text{or }}x = {e^{\frac{{i\pi }}{3}}} \cr
& \therefore {x^a} + {x^{ - a}} = {e^{\frac{{ia\pi }}{3}}} + {e^{\frac{{ - ia}}{3}}} = 2\cos \frac{{ar}}{3} \cr
& {\text{Hence, }}\cos \frac{{a\pi }}{3} + \cos \frac{{b\pi }}{3} + \cos \frac{{c\pi }}{3} = 0 \cr
& a,b,c \in I \cr
& \therefore {\left. {a + b + c} \right|_{\min }} = \left( {1 + 3 + 5} \right) = 9 \cr} $$