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To solve this problem, let’s break down each part and calculate the required values using differentials. 

Given:
- Edge length of the cube, \( x = 15 \) cm.
- Possible error in measurement, \( dx = 0.1 \) cm.

We need to find:
1. The maximum possible error, relative error, and percentage error in computing:
   - (a) the volume of the cube.
   - (b) the surface area of the cube.

---

### (a) Volume of the Cube

The volume \( V \) of a cube with edge length \( x \) is given by:
\[
V = x^3
\]

#### 1. Differential of Volume
To estimate the maximum possible error in the volume, we use the differential:
\[
dV = 3x^2 \cdot dx
\]

Plugging in the values:
\[
dV = 3 \times (15)^2 \times 0.1 = 3 \times 225 \times 0.1 = 67.5 \, \text{cm}^3
\]

So, the **maximum possible error in volume** is \( 67.5 \, \text{cm}^3 \).

#### 2. Relative Error in Volume
The relative error is given by:
\[
\text{Relative Error} = \frac{dV}{V}
\]

Calculating \( V \) with \( x = 15 \):
\[
V = 15^3 = 3375 \, \text{cm}^3
\]

Then,
\[
\text{Relative Error} = \frac{67.5}{3375} \approx 0.02
\]

So, the **relative error in volume** (rounded to four decimal places) is \( 0.0200 \).

#### 3. Percentage Error in Volume
The percentage error is simply the relative error multiplied by 100:
\[
\text{Percentage Error}

The edge of a cube was found to be 15 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube.

Learn how to estimate the maximum possible error, relative error, and percentage error in the volume and surface area of a cube with a 15 cm edge and a 0.1 cm measurement error using differentials. This solution involves key calculus concepts, including error analysis and differentiation.

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