(d) 0
For $x^2 - 1$, we have $a = 1,\ b = 0,\ c = -1$. So $\alpha + \beta = \dfrac{-b}{a} = \dfrac{0}{1} = 0$.
Source: Chapter 2, Section 2.3
Use the formula $\alpha + \beta = \frac{-b}{a}$. Since the polynomial $x^2 - 1$ has no $x$ term, $b = 0$, making the sum of zeroes zero. You can verify: zeroes are $+1$ and $-1$, and $1 + (-1) = 0$.