Finding the value of an expression using substitution

Question

How to find the value of $x^{4000} + \frac{1}{x^{4000}}$ if $x+\frac{1}{x}=1$.
I think binomial theorem will be useful in this.

Here's my proceedings: $$x+\frac{1}{x}=1$$ Raising both sides to the power of 4000 $$(x+\frac{1}{x})^{4000} = 1^{4000}$$ $${4000 \choose 0} x^{4000} + {4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... +{4000 \choose 4000} \frac{1}{x^{4000}} = 1$$ $$x^{4000} + \frac{1}{x^{4000}} = 1 - ({4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... + {4000 \choose 3999} x \frac{1}{x^{3999}})$$ $$x^{4000} + \frac{1}{x^{4000}} = 1- \sum\limits_{r=1}^{3999} {4000 \choose r} x^{4000-r} \frac{1}{x^r}$$ Now how to proceed further to get to my answer?

Answer 1

HINT:

$$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\implies x^3=-1$$

Now $4000=3(2n+1)+1$

Now $x^{3(2n+1))}=(x^3)^{2n+1}\cdot x=(-1)^{2n+1}\cdot x=-x$

Answer 2

Hint: Our number $x$ is a primitive sixth root of unity. Taking high powers of it is exceptionally easy.