If $ x^2+x+1=0$ find the value of $8x^{282}+1799x^{183}+87x^{51}+124x^{-3}+1$

Question

If$$ x^2+x+1=0$$ find the value of $$8x^{282}+1799x^{183}+87x^{51}+124x^{-3}+1$$

Solving this equation gives imaginary solutions.

Is there an easy way to do this ?

Answer 1

Hint:

Observe that such a $\;x\;$ fufills $\;x^3=1\;$ ...

Answer 2

With $x^3-1=(x-1)(x^2+x+1)$ we can use $x^{3}=1$ and then reduce the polynomial to $f(x)=8+1799+87+124+1=2019$. This is almost $3$ years ahead, though.