1. **Problem statement:** Given the cubic equation $x^3 - 8x - 1 = 0$, it has a real root between two consecutive integers $n$ and $n+1$. The function values $f(x) = x^3 - 8x - 1$ at integers $x$ from $-3$ to $3$ are given as follows: | x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | |---|----|----|----|---|---|---|---| | f(x) | -4 | 7 | 6 | -1 | -8 | -9 | 2 | Find the sum of all possible values of $n$ where the root lies between $n$ and $n+1$. 2. **Formula and rule:** A root exists between two consecutive integers $n$ and $n+1$ if the function values at these points have opposite signs, i.e., if $f(n) imes f(n+1) < 0$. 3. **Check intervals:** - Between $-3$ and $-2$: $f(-3) = -4$, $f(-2) = 7$, product $-4 imes 7 = -28 < 0$ (root exists) - Between $-2$ and $-1$: $7 imes 6 = 42 > 0$ (no root) - Between $-1$ and $0$: $6 imes (-1) = -6 < 0$ (root exists) - Between $0$ and $1$: $-1 imes (-8) = 8 > 0$ (no root) - Between $1$ and $2$: $-8 imes (-9) = 72 > 0$ (no root) - Between $2$ and $3$: $-9 imes 2 = -18 < 0$ (root exists) 4. **Possible $n$ values:** $-3$, $-1$, and $2$. 5. **Sum of all possible $n$ values:** $$-3 + (-1) + 2 = -2$$ **Final answer:** (B) -2