Find $a^{100}+b^{100}+ab$ $a$ and $b$ are the roots of the equation $x^2+x+1=0$.
Then what is the value of $a^{100}+b^{100}+ab$?
Here's what I found out:
$$a+b=-1$$
$$ab=1$$
but how to use this to find that I don't know! Someone please answer my query.

The roots of your equation are $$x = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$$
We have $e^{\frac{2\pi i}{3}}$ and $e^{\frac{4\pi i}{3}}$, both which remain unchanged in magnitude and direction upon exponentiating 100 times as $e^{\frac{2\pi i}{3}} = e^{\frac{200 \pi i}{3}}$ and $e^{\frac{4\pi i}{3}} = e^{\frac{400 \pi i}{3}}$, telling us that $a^{100} = a$ and $b^{100} = b$ and thus $a + b + ab = 0$.
