Let $r_i$ (for $i = 1$, $2$, $\dots$, $7$) be distinct roots in $\mathbb C$ of the polynomial $f(x) = x^7 - 7$ and let $$M = \prod_{1\leq i < j \leq 7} (r_i + r_j)$$ The problem is to compute the value of $M^2$.
Given the above distinct roots, we can rewrite the polynomial by $$\prod_{i=1}^7(x-r_i) = x^7 - 7$$ Define $g(x) = - f(-x)$ so that $$g(x) = \prod_{i=1}^7(x+r_i) = x^7 + 7$$ Clearly, $g(0) = 7$ implies $\prod\limits_{i=1}^7 r_i = 7$. There has to be a way to use this fact to obtain $M$, but I don't know what further must be done. The correct answer is supposed to be $M^2 = 7^6$.
Hint: $\displaystyle \;M^2 = \prod_{1 \le i \ne j \le 7} (r_i+r_j)=\frac{1}{\prod_{1 \le i \le 7}(r_i+r_i)} \prod_{1\le i,j \le 7} (r_i+r_j) = \frac{1}{2^7 \cdot 7} \prod_{1 \le i \le 7} g(r_i)\,$.