If `m=root3(125)` and `n=root3(64)` , find the value of `m-n-(1)/(m^(2)+mn+n^(2))`.

Answer

To solve the problem, we need to find the value of the expression \( m - n - \frac{1}{m^2 + mn + n^2} \) where \( m = \sqrt[3]{125} \) and \( n = \sqrt[3]{64} \).

### Step-by-Step Solution:

1. **Calculate \( m \) and \( n \)**:
- \( m = \sqrt[3]{125} = \sqrt[3]{5^3} = 5 \)
- \( n = \sqrt[3]{64} = \sqrt[3]{4^3} = 4 \)

2. **Find \( m - n \)**:
- \( m - n = 5 - 4 = 1 \)

3. **Calculate \( m^2 + mn + n^2 \)**:
- First, calculate \( m^2 \):
\[ m^2 = 5^2 = 25 \]
- Next, calculate \( mn \):
\[ mn = 5 \times 4 = 20 \]
- Then, calculate \( n^2 \):
\[ n^2 = 4^2 = 16 \]
- Now, sum them up:
\[ m^2 + mn + n^2 = 25 + 20 + 16 = 61 \]

4. **Substitute values into the expression**:
- Now we substitute \( m - n \) and \( m^2 + mn + n^2 \) into the expression:
\[ m - n - \frac{1}{m^2 + mn + n^2} = 1 - \frac{1}{61} \]

5. **Simplify the expression**:
- To simplify \( 1 - \frac{1}{61} \):
\[ 1 = \frac{61}{61} \]
Thus,
\[ 1 - \frac{1}{61} = \frac{61}{61} - \frac{1}{61} = \frac{60}{61} \]

### Final Answer:
The value of \( m - n - \frac{1}{m^2 + mn + n^2} \) is \( \frac{60}{61} \).

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