If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is? 
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?

Ans. 4
What I have  tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^{3}
\end{align}
as $1 + a^{2} = -a$ and $1 + a = -a^{2}$ from $(1)$.

Without solving the quadratic:
$$ a^3 = - a^2 - a = a + 1 - a = 1 $$
which was found by using the equation $a^2 + a + 1 = 0$ twice. This means that $a$ is a non-real cube root of unity.
