<s>[INST] 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 $(1)$.
 [/INST] Just solve the quadratic $(1)$ to get $a$, then substitute $a$ back in the original expression or $4a^3$.
Also, since $a^3-1=(a-1)(a^2+a+1),$ $a^3=1$.
In more detail, from $(1)$,
$$\begin{align}a&=\frac{-1\pm\sqrt{1-4}}{2} \\
&=\frac{-1\pm i\sqrt{3}}{2}.\end{align}$$
Now
$$\begin{align}
a^2&=\frac{-1\pm i\sqrt{3}}{2}\times \frac{-1\pm i\sqrt{3}}{2} \\
&=\frac{1\mp i2\sqrt{3}-3}{4} \\
&=\frac{-1\mp i\sqrt{3}}{2}
\end{align}$$ and 
$$\begin{align}
a^3&=\frac{-1\pm i\sqrt{3}}{2}\times \frac{-1\mp i\sqrt{3}}{2} \\
&=\frac{1\pm i\sqrt{3}\mp i\sqrt{3}+3}{4} \\
&=1.
\end{align}$$
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