In the following find the approximate values, using differentials :
`(0.731)^(1//3)`

Answer

To approximate the value of \( (0.731)^{1/3} \) using differentials, we can follow these steps:

### Step 1: Identify a suitable function and point
Let \( y = x^{1/3} \). We want to approximate \( y \) at \( x = 0.731 \). We can choose a nearby point where we know the exact value. The closest perfect cube is \( 0.729 \) (which is \( 0.9^3 \)), so we will use \( x = 0.729 \).

### Step 2: Calculate \( \Delta x \)
We can define \( \Delta x \) as the difference between our target value and the known value:
\[ \Delta x = 0.731 - 0.729 = 0.002 \]

### Step 3: Find \( y \) at the known point
Now, we calculate \( y \) at \( x = 0.729 \):
\[ y = (0.729)^{1/3} = 0.9 \]

### Step 4: Differentiate the function
Next, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{3} x^{-2/3} \]

### Step 5: Evaluate the derivative at the known point
Now we evaluate the derivative at \( x = 0.729 \):
\[ \frac{dy}{dx} \bigg|_{x=0.729} = \frac{1}{3} (0.729)^{-2/3} \]
Calculating \( (0.729)^{-2/3} \):
\[ (0.729)^{1/3} = 0.9 \quad \text{so} \quad (0.729)^{-2/3} = \frac{1}{(0.9)^2} = \frac{1}{0.81} \]
Thus,
\[ \frac{dy}{dx} \bigg|_{x=0.729} = \frac{1}{3} \cdot \frac{1}{0.81} = \frac{1}{2.43} \]

### Step 6: Calculate \( \Delta y \)
Now we can find \( \Delta y \) using the formula:
\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]
Substituting the values we have:
\[ \Delta y \approx \frac{1}{2.43} \cdot 0.002 \]
Calculating \( \Delta y \):
\[ \Delta y \approx \frac{0.002}{2.43} \approx 0.00082 \]

### Step 7: Find the approximate value of \( (0.731)^{1/3} \)
Finally, we can find the approximate value of \( (0.731)^{1/3} \):
\[ (0.731)^{1/3} \approx 0.9 + 0.00082 \approx 0.90082 \]

### Final Answer
Thus, the approximate value of \( (0.731)^{1/3} \) is:
\[ \boxed{0.90082} \]