Find the cube root of 217, correct to two decimal places.

Explanation

To find the cube root of 217 correct to two decimal places using the Binomial Theorem, we can follow these steps:

### Step 1: Rewrite the number
We start by expressing 217 in a form that is easier to work with. We can write:
\[ 217 = 6^3 + 1 \]
This means we can express the cube root as:
\[ \sqrt[3]{217} = \sqrt[3]{6^3 + 1} \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we can express the cube root as:
\[ \sqrt[3]{6^3 + 1} = 6 \cdot \sqrt[3]{1 + \frac{1}{6^3}} \]
Here, \(x = \frac{1}{6^3} = \frac{1}{216}\).

### Step 3: Use the Binomial Expansion
According to the Binomial Theorem, for small \(x\):
\[ (1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \ldots \]
For our case, \(n = \frac{1}{3}\) and \(x = \frac{1}{216}\):
\[ \sqrt[3]{1 + \frac{1}{216}} \approx 1 + \frac{1}{3} \cdot \frac{1}{216} + \frac{\frac{1}{3} \cdot \left(\frac{1}{3} - 1\right)}{2} \cdot \left(\frac{1}{216}\right)^2 \]

### Step 4: Calculate the first term
Calculating the first term:
\[ 1 + \frac{1}{3} \cdot \frac{1}{216} = 1 + \frac{1}{648} \]

### Step 5: Calculate the second term
Now calculating the second term:
\[ \frac{\frac{1}{3} \cdot \left(-\frac{2}{3}\right)}{2} \cdot \left(\frac{1}{216}\right)^2 = -\frac{1}{9} \cdot \frac{1}{46656} = -\frac{1}{419904} \]
Since this term is very small, we can neglect it for our approximation.

### Step 6: Combine the results
Now we can combine our results:
\[ \sqrt[3]{1 + \frac{1}{216}} \approx 1 + \frac{1}{648} \]
Thus:
\[ \sqrt[3]{217} \approx 6 \cdot \left(1 + \frac{1}{648}\right) = 6 + \frac{6}{648} = 6 + \frac{1}{108} \approx 6 + 0.00926 \approx 6.00926 \]

### Step 7: Round to two decimal places
Finally, rounding to two decimal places, we get:
\[ \sqrt[3]{217} \approx 6.01 \]

### Final Answer
The cube root of 217, correct to two decimal places, is:
\[ \boxed{6.01} \]