How would you find the exact roots of $y=x^3+x^2-2x-1$?

Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$

The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$ ( coming from the $t^7 - 1$ factor in numerator ) for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\frac{2k\pi}{7} i} = 2\cos\left(\frac{2 k\pi}{7}\right)$$ for $k = 1,2,3$.

Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$

Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$ $$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$

Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated in your question. Note that $$2\cos\frac{6\pi}{7}=2\cos\frac{8\pi}{7}$$

Set $x=t+t^{-1}$. Then the equation becomes $$ t^3+3t+3t^{-1}+t^{-3}+t^2+2+t^{-2}-2t-2t^{-1}-1=0 $$ and, multiplying by $t^3$, $$ t^6+t^5+t^4+t^3+t^2+t+1=0 $$ and it should be now clear what the solutions are. For each root there's another one giving the same solution in $x$.