[b]i.)[/b] The polynomial $x^{2 \cdot k} + 1 + (x+1)^{2 \cdot k}$ is not divisible by $x^2 + x + 1.$ Find the value of $k.$

[b]ii.)[/b] If $p,q$ and $r$ are distinct roots of $x^3 - x^2 + x - 2 = 0$ the find the value of $p^3 + q^3 + r^3.$

[b]iii.)[/b] If $r$ is the remainder when each of the numbers 1059, 1417 and 2312 is divided by $d,$ where $d$ is an integer greater than one, then find the value of $d-r.$

[b]iv.)[/b] What is the smallest positive odd integer $n$ such that the product of

\[ 2^{\frac{1}{7}}, 2^{\frac{3}{7}}, \ldots, 2^{\frac{2 \cdot n + 1}{7}} \]

is greater than 1000?
### Part i)
Given the polynomial \( f(x) = x^{2k} + 1 + (x+1)^{2k} \), we need to determine the value of \( k \) such that \( f(x) \) is not divisible by \( x^2 + x + 1 \).

1. The roots of \( x^2 + x + 1 \) are the complex cube roots of unity, \( \omega \) and \( \omega^2 \), where \( \omega = e^{i\frac{2\pi}{3}} \) and \( \omega^3 = 1 \).

2. Evaluate \( f(x) \) at \( x = \omega \):
   \[
   f(\omega) = \omega^{2k} + 1 + (\omega + 1)^{2k}
   \]
   Since \( \omega^3 = 1 \), we have \( \omega^2 = \omega^{-1} \) and \( \omega + 1 = -\omega^2 \). Thus,
   \[
   f(\omega) = \omega^{2k} + 1 + (-\omega^2)^{2k} = \omega^{2k} + 1 + \omega^{4k}
   \]

3. For \( f(x) \) to be divisible by \( x^2 + x + 1 \), \( f(\omega) \) must be zero:
   \[
   \omega^{2k} + 1 + \omega^{4k} = 0
   \]

4. Since \( \omega^3 = 1 \), we can write \( \omega^{4k} = \omega^{4k \mod 3} \). Therefore, the equation becomes:
   \[
   \omega^{2k} + 1 + \omega^{4k \mod 3} = 0
   \]

5. For \( \omega^{2k} + 1 + \omega^{4k \mod 3} = 0 \) to hold, \( 2k \mod 3 \) and \( 4k \mod 3 \) must satisfy the equation. This implies that \( k \) must be a multiple of 3.

6. If \( k \) is not a multiple of 3, then \( f(\omega) \neq 0 \), and thus \( f(x) \) is not divisible by \( x^2 + x + 1 \).

Therefore, the value of \( k \) such that \( f(x) \) is not divisible by \( x^2 + x + 1 \) is when \( k \) is not a multiple of 3.

### Part ii)
Given the polynomial \( x^3 - x^2 + x - 2 = 0 \) with roots \( p, q, r \), we need to find the value of \( p^3 + q^3 + r^3 \).

1. By Vieta's formulas, we know:
   \[
   p + q + r = 1, \quad pq + qr + rp = 1, \quad pqr = 2
   \]

2. Using the identity for the sum of cubes:
   \[
   p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2 + q^2 + r^2 - pq - qr - rp)
   \]

3. Substitute the known values:
   \[
   p^3 + q^3 + r^3 - 3 \cdot 2 = 1 \cdot (p^2 + q^2 + r^2 - pq - qr - rp)
   \]

4. Simplify \( p^2 + q^2 + r^2 \) using the square of the sum of the roots:
   \[
   (p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp)
   \]
   \[
   1^2 = p^2 + q^2 + r^2 + 2 \cdot 1
   \]
   \[
   1 = p^2 + q^2 + r^2 + 2
   \]
   \[
   p^2 + q^2 + r^2 = -1
   \]

5. Substitute back into the identity:
   \[
   p^3 + q^3 + r^3 - 6 = 1 \cdot (-1 - 1)
   \]
   \[
   p^3 + q^3 + r^3 - 6 = -2
   \]
   \[
   p^3 + q^3 + r^3 = 4
   \]

### Part iii)
Given the numbers 1059, 1417, and 2312, we need to find the value of \( d - r \), where \( r \) is the remainder when each number is divided by \( d \).

1. The condition \( 1059 \equiv 1417 \equiv 2312 \pmod{d} \) implies:
   \[
   d \mid (1417 - 1059), \quad d \mid (2312 - 1417), \quad d \mid (2312 - 1059)
   \]

2. Calculate the differences:
   \[
   1417 - 1059 = 358, \quad 2312 - 1417 = 895, \quad 2312 - 1059 = 1253
   \]

3. Find the greatest common divisor (GCD) of 358, 895, and 1253:
   \[
   \gcd(358, 895, 1253) = 179
   \]

4. Therefore, \( d = 179 \).

5. Find the common remainder \( r \):
   \[
   1059 \equiv 1417 \equiv 2312 \equiv 164 \pmod{179}
   \]

6. Calculate \( d - r \):
   \[
   d - r = 179 - 164 = 15
   \]

### Part iv)
Given the product \( 2^{\frac{1}{7}}, 2^{\frac{3}{7}}, \ldots, 2^{\frac{2n+1}{7}} \), we need to find the smallest positive odd integer \( n \) such that the product is greater than 1000.

1. The product can be written as:
   \[
   \prod_{k=0}^{n} 2^{\frac{2k+1}{7}} = 2^{\sum_{k=0}^{n} \frac{2k+1}{7}}
   \]

2. Simplify the exponent:
   \[
   \sum_{k=0}^{n} \frac{2k+1}{7} = \frac{1}{7} \sum_{k=0}^{n} (2k+1)
   \]

3. The sum of the first \( n+1 \) odd numbers is:
   \[
   \sum_{k=0}^{n} (2k+1) = (n+1)^2
   \]

4. Therefore, the exponent becomes:
   \[
   \frac{(n+1)^2}{7}
   \]

5. We need this to be greater than \( \log_2(1000) \):
   \[
   2^{\frac{(n+1)^2}{7}} > 1000
   \]
   \[
   \frac{(n+1)^2}{7} > \log_2(1000)
   \]
   \[
   \log_2(1000) \approx 10
   \]
   \[
   \frac{(n+1)^2}{7} > 10
   \]
   \[
   (n+1)^2 > 70
   \]
   \[
   n+1 > \sqrt{70} \approx 8.37
   \]
   \[
   n > 7.37
   \]

6. The smallest positive odd integer \( n \) is 9.

The final answer is \( \boxed{ k } \) is not a multiple of 3, \( p^3 + q^3 + r^3 = 4 \), \( d - r = 15 \), and \( n = 9 \).