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.

Here is a solution that does not require flashes of inspiration.
$a^2+a+1=0$ implies $a^2=-a-1$ and so $a^3=-a(a+1)=-(a^2+a)=-(-a-1+a)=1$. Therefore, $a^{100}=a^{99}a=(a^3)^{33}a=a$.
The same holds for $b$ and so
$$
a^{100}+b^{100}+ab=a+b+ab=-1+1=0
$$
