Is this a real formula? If not, is it a valid statement?
$$\begin{align} r_1^3 + r_2^3 + r_3^3 + r_4^3 + r_5^3 + r_6^3 + r_7^3 &= (r_1 + r_2 + r_3 + r_4 + r_5 + r_6 + r_7)^3 \\ &- 3(r_1 + r_2 + \cdots + r_7)(r_1r_2 + r_1r_3 + \cdots + r_6r_7) \\ &+ 3(r_1r_2r_3 + r_1r_2r_4 + \cdots + r_5r_6r_7) \end{align}$$
If this is real, is there a general formula for this that's not JUST $n = 7$? It can be any integer?
I'm trying to use this identity to solve this problem:
The polynomial $x^7 + x^6 + x^4 + x^3 + x + 1$ has roots $r_1,$ $r_2,$ $\dots,$ $r_7.$ Calculate $$\sum_{n = 1}^7 \left( r_n^3 + \frac{1}{r_n^3} \right).$$
So far, I have used the vieta's formula and I have made decent progress if that formula I stated above is true.
I'm stuck on another issue, however:
How to apply vieta's formula to this?
$$\frac {1} {r_1^3} + \frac {1} {r_2^3} + \frac {1} {r_3^3} + \frac {1} {r_4^3} + \frac {1} {r_5^3} + \frac {1} {r_6^3} + \frac {1} {r_7^3}$$
You already have a route using Newton's identities. Here is just another way, perhaps simpler in this case: $$x^7+x^6+x^4+x^3+x+1= (x+1)(x^6+x^3+1)=0$$ $$\implies (x+1)\left(x^3+\frac1{x^3}+1 \right)=0$$ So, apart from the root $x=-1$ say $r_1$, the rest of the roots satisfy $x^3+\dfrac1{x^3}=-1$. Hence $$\sum_{n=1}^7\left(r_n^3+\frac1{r_n^3} \right)=-2+6\cdot(-1)=-8$$