If  ω  be the non real cube root of unity &  z  =  r=2∑8(r − 1)(r − w)(r − w2)  then

1. Break Down the Problem

Given the problem involves the non-real cube roots of unity and a sum with a polynomial, we'll approach this by first understanding the roots of unity, specifically the values of $\omega$ and $\omega^2$.

2. Relevant Concepts

The non-real cube roots of unity are defined as:

• $\omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$
• $\omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$

The properties of the roots include:

• $\omega^3 = 1$
• $1 + \omega + \omega^2 = 0$

The expression to evaluate is $z = r = 2 \sum_{r=1}^{8}(r - 1)(r - \omega)(r - \omega^2)$.

3. Analysis and Detail

To simplify the sum, we expand the polynomial term:

$(r - 1)(r - \omega)(r - \omega^2)$

This can be expanded as follows: $(r - 1)((r - \omega)(r - \omega^2)) = (r - 1)(r^2 - (\omega + \omega^2)r + \omega\omega^2) = (r - 1)(r^2 + r - 1)$

Where we used $\omega + \omega^2 = -1$ and $\omega \cdot \omega^2 = 1$.

Thus, we need to calculate: $\sum_{r=1}^{8} (r - 1)(r^2 + r - 1)$

Evaluating this sum separately for $r$ from 1 to 8 and then multiplying by 2.

4. Verify and Summarize

First, let's find the expressions:

$(r - 1)(r^2 + r - 1) = (r-1)(r^2 + r - 1) = r^3 + r^2 - r - (r^2 + r - 1) = r^3 - 1$

Now calculate the sum: $\sum_{r=1}^{8} (r^3 - 1)$

Using the formula for the sum of cubes, we have: $\sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2$ For $n = 8$: $\sum_{r=1}^{8} r^3 = \left(\frac{8 \cdot 9}{2}\right)^2 = 36^2 = 1296$ Thus, $\sum_{r=1}^{8} r^3 - 8 = 1296 - 8 = 1288$

Final calculation for $z$: $z = 2 \times 1288 = 2576$

Final Answer

The value of $z$ is $2576$.