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### Problem 1:
- **Given:**
  The area of a rectangular flower bed is 63 m².
  
- **Find:**
  The width of the path around the flower bed.

We do not have enough information to determine the exact method to solve this (it likely requires additional dimensions like length or width of the flower bed), but let's move forward with solving the next problems for now.

### Problem 2:
- **Given:**
  There is a square flower bed with a side length of \( x \). When a 3-meter wide path is added around it, the remaining flower bed area becomes 70 m².
  
- **Find:**
  The original side length \( x \) of the square.

Let's solve this algebraically:
1. The total area of the flower bed before the path was added is \( x^2 \).
2. The width of the path is 3 meters. After subtracting the area taken by the path, the remaining area of the flower bed is given to be 70 m².
3. The new side length after subtracting the path is \( x - 6 \) (since the path takes 3 meters from each side).

Thus, we can set up the equation:
\[
(x - 6)^2 = 70
\]
Let's solve this equation for \( x \).

### Problem 3:
- **Given:**
  There is a square piece of paper. When a 4 cm square is cut from each corner, a box is formed with a volume of 256 cm³.

- **Find:**
  The side length of the original square paper.

Let's solve this step by step:
1. Let the side length of the original square paper be \( y \).
2. After cutting the 4 cm squares from each corner, the height of the box becomes 4 cm, and the new base dimensions are \( (y - 8) \times (y - 8) \), since 4 cm is removed from both sides.
3. The volume of the box is given by:
\[
(y - 8)^2 \times 4 = 256
\]
Let's solve this equation for \( y \).

I'll now compute the results for Problems 2 and 3.Here are the solutions:

### Problem 2:
The equation \((x - 6)^2 = 70\) yields two solutions:
\[
x = 6 - \sqrt{70} \quad \text{or} \quad x = 6 + \sqrt{70}
\]
Since \( x \) represents a length, the positive solution is the valid one:
\[
x = 6 + \sqrt{70} \approx 14.37 \, \text{m}
\]
Thus, the original side length of the square flower bed is approximately **14.37 meters**.

### Problem 3:
The equation \((y - 8)^2 \times 4 = 256\) yields two solutions:
\[
y = 0 \quad \text{or} \quad y = 16
\]
Since \( y \) represents the side length of the paper, the valid solution is:
\[
y = 16 \, \text{cm}
\]
Thus, the original side length of the square paper is **16 cm**.

Would you like more details or have any further questions?

Here are five related questions to further expand your understanding:
1. How would the solution change if the path in Problem 2 were 4 meters wide instead of 3 meters?
2. Can you calculate the perimeter of the flower bed after adding the path in Problem 2?
3. What would happen to the volume of the box in Problem 3 if the cut squares were 5 cm instead of 4 cm?
4. How does the total area of the paper compare before and after the cuts in Problem 3?
5. How can you verify the solution to Problem 3 using geometric reasoning?

**Tip:** When solving quadratic equations, always remember to discard non-physical solutions that don't make sense in the context of the problem (e.g., negative lengths).

The original question submitted by the user as text.

This lesson explains how to solve quadratic equations and volume problems. We calculate the original side length of a square flower bed and a square piece of paper cut to form a box with a given volume. Suitable for students in Grades 7-9.

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