Let n be an integer greater than 7. Let $z^n-1=0$ has roots $1,\lambda_1,\lambda_2,...,\lambda_{n-1}$ then
(a) $\sum_{i=1}^{n-1}\lambda_i=-1$
(b) $\sum\sum_{1\le i <j \le n-1}\lambda_i\lambda_j=1$
(c) $\sum\sum\sum_{1\le i <j <k\le n-1}\lambda_i\lambda_j\lambda_k=-1$
(d) $\lambda_1\lambda_2...\lambda_{n-1}=(-1)^n$
Options (a) and (d) are fairly standard results and squaring (a) I got the result in option (b). I am clueless about option (c) which seems true but I am unable to prove it as yet, for which I seek your help
As ancient mathematician suggested, setting $\lambda_0 = 1$ is a good idea. Vieta's formula (which is essentially just writing the polynomial as a product of its linear factors) then shows $$\sum\limits_{0 \leq i < j < k \leq n-1} \lambda_i \lambda_j \lambda_k = 0.$$ Since $$\sum\limits_{0 \leq i < j < k \leq n-1} \lambda_i \lambda_j \lambda_k = \sum\limits_{1 \leq i < j < k \leq n-1} \lambda_i \lambda_j \lambda_k + \sum\limits_{1 \leq j < k \leq n-1} \lambda_0 \lambda_j \lambda_k,$$ this means $\sum\limits_{1 \leq i < j < k \leq n-1} \lambda_i \lambda_j \lambda_k = -1$ by (b).
Similarly you can see that for the sum of all products with $m$ different (non-trivial) roots you get $(-1)^{m-1}$ as a result.