Use Euclid’s division lemma to show that the cube of any positive integer is of the form 7m, 7m+1, 7m+6.

To solve this problem using Euclid's division lemma, let's proceed step by step.

### Euclid's Division Lemma:
According to Euclid's division lemma, for any two integers \( a \) and \( b \), where \( b > 0 \), there exist unique integers \( q \) (quotient) and \( r \) (remainder) such that:
\[ a = bq + r \]
where \( 0 \leq r < b \).

Here, we are working with the divisor \( b = 7 \), so any integer \( n \) can be written as:
\[ n = 7q + r \]
where \( r \) can take values \( 0, 1, 2, 3, 4, 5, 6 \).

### Step 1: Generalize the cube of \( n \)
The cube of \( n \) is:
\[ n^3 = (7q + r)^3 \]

### Step 2: Analyze cases for \( r \)
We will expand \( (7q + r)^3 \) for each possible value of \( r \).

1. **Case \( r = 0 \):**
\[ n = 7q + 0 \]
\[ n^3 = (7q)^3 = 343q^3 \]
Clearly, \( n^3 \) is a multiple of \( 7 \), i.e., \( n^3 = 7m \), where \( m = 49q^3 \).

2. **Case \( r = 1 \):**
\[ n = 7q + 1 \]
\[ n^3 = (7q + 1)^3 = 343q^3 + 147q^2 + 21q + 1 \]
\[ n^3 = 7(49q^3 + 21q^2 + 3q) + 1 \]
Thus, \( n^3 = 7m + 1 \), where \( m = 49q^3 + 21q^2 + 3q \).

3. **Case \( r = 2 \):**
\[ n = 7q + 2 \]
\[ n^3 = (7q + 2)^3 = 343q^3 + 294q^2 + 84q + 8 \]
\[ n^3 = 7(49q^3 + 42q^2 + 12q) + 8 \]
Since \( 8 \mod 7 = 1 \), this simplifies to:
\[ n^3 = 7m + 1 \], where \( m = 49q^3 + 42q^2 + 12q \).

4. **Case \( r = 3 \):**
\[ n = 7q + 3 \]
\[ n^3 = (7q + 3)^3 = 343q^3 + 441q^2 + 147q + 27 \]
\[ n^3 = 7(49q^3 + 63q^2 + 21q) + 27 \]
Since \( 27 \mod 7 = 6 \), this simplifies to:
\[ n^3 = 7m + 6 \], where \( m = 49q^3 + 63q^2 + 21q \).

5. **Case \( r = 4 \):**
\[ n = 7q + 4 \]
\[ n^3 = (7q + 4)^3 = 343q^3 + 588q^2 + 196q + 64 \]
\[ n^3 = 7(49q^3 + 84q^2 + 28q) + 64 \]
Since \( 64 \mod 7 = 1 \), this simplifies to:
\[ n^3 = 7m + 1 \], where \( m = 49q^3 + 84q^2 + 28q \).

6. **Case \( r = 5 \):**
\[ n = 7q + 5 \]
\[ n^3 = (7q + 5)^3 = 343q^3 + 735q^2 + 245q + 125 \]
\[ n^3 = 7(49q^3 + 105q^2 + 35q) + 125 \]
Since \( 125 \mod 7 = 6 \), this simplifies to:
\[ n^3 = 7m + 6 \], where \( m = 49q^3 + 105q^2 + 35q \).

7. **Case \( r = 6 \):**
\[ n = 7q + 6 \]
\[ n^3 = (7q + 6)^3 = 343q^3 + 882q^2 + 294q + 216 \]
\[ n^3 = 7(49q^3 + 126q^2 + 42q) + 216 \]
Since \( 216 \mod 7 = 6 \), this simplifies to:
\[ n^3 = 7m + 6 \], where \( m = 49q^3 + 126q^2 + 42q \).

### Step 3: Conclusion
From all the cases above, the cube of any positive integer \( n \) can only take one of the forms \( 7m, 7m + 1, 7m + 6 \), depending on the remainder \( r \) when \( n \) is divided by \( 7 \).