If $x$ +$\frac{1}{x}$ = $\sqrt{3}$, then find the value of $x^{18}$ + $x^{12}$ + $x^6$ +1.
I think there is a more ingenious way to solve this.....but....by brute force ......
x + 1/x = sqrt (3)
x^2 + 1 = sqrt (3) x
x^2 - sqrt (3) x = -1
x^2 - sqrt (3) x + 3/4 = -1 + 3/4
(x + sqrt (3) /2) ^2 = -1/4 {take the positive root since x + 1/x is positive}
x - sqrt (3)/2 = (1/2) i
x = sqrt (3/2) + (1/2)i
x^3 = (sqrt (3)/2 + (1/2) i)^3 =
((1/2) (sqrt (3) + i)^3 {binomial expansion of terms in the parentheses }
(1/8) ( 3sqrt (3) + 3[sqrt (3)]^2 i + 3sqrt (3)i^2 + i^3) =
(1/8) ( 3sqrt (3) + 9i - 3sqr (3) - i) =
(1/8) ( 8i) = i
i = x
i^18 = i^2 = -1
i^12 = 1
i^6 = -1
So
i^18 + i^12 + i^6 + 1 =
-1 + 1 - 1 + 1 = 0