Find $a^{100}+b^{100}+ab$

Question

$a$ 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.

Answer 1

Let $\omega$ be a root of $x^2+x+1=0$, then $\omega^3=1$. Thus $$ a^{100}+b^{100}+ab=a+b+ab. $$

Answer 2

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 $a + b + ab = 0$.

Answer 3

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 $$ a^{100}+b^{100}+ab=a+b+ab=-1+1=0 $$