You are told that in a billiards shot, the cue ball was shot at the eight ball, which was 8 inches away. As a result, the eight ball rolled into a pocket, which was 6 inches away. Knowing that the angle made with the path of the cue ball and the resulting path of the eight ball is larger than $$ 90^{\circ} $$, it can be determined that the original distance from the cue ball to the pocket was greater than $$ \square $$

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To determine the original distance from the cue ball to the pocket, let's analyze the given information using geometry.

### Given:
1. **Distance between the cue ball and the eight ball (CO):** 8 inches.
2. **Distance between the eight ball and the pocket (OP):** 6 inches.
3. **Angle between the path of the cue ball and the path of the eight ball (∠COP):** Greater than $$90^\circ$$.

### Objective:
Find the original distance from the cue ball (C) to the pocket (P), denoted as $$CP$$.

### Analysis:
We can model the situation using the Law of Cosines in triangle $$COP$$:

$$
CP^2 = CO^2 + OP^2 - 2 \cdot CO \cdot OP \cdot \cos(\angle COP)
$$

Given that $$\angle COP > 90^\circ$$, the cosine of this angle is negative ($$\cos(\angle COP) < 0$$). This implies:

$$
CP^2 > CO^2 + OP^2
$$

Plugging in the known values:

$$
CP^2 > 8^2 + 6^2 \
CP^2 > 64 + 36 \
CP^2 > 100 \
CP > \sqrt{100} \
CP > 10 \text{ inches}
$$

### Conclusion:
The original distance from the cue ball to the pocket must be **greater than 10 inches**.

**Answer:**  
Its original distance must have been larger than ten inches.
The original distance from the cue ball to the pocket was greater than 10 inches.

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