For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$ Consider the following problem:

Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities:
$$
\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2},
\\
\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}.
$$

Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities
$$
(a+b)^3-a^3-b^3=3ab(a+b),
\\
(a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2),
\\
(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2.
$$
However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.

Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.

Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.

Here's another way:
Let $a,b,c$ be the roots of the cubic equation $$x^3+px+q=0$$
(The $x^2$ coefficient is zero since $\sum a=0$)
Then $$\sum a^2=(\sum a)^2-2\sum ab=-2p$$
$$\sum a^3=-p\sum a-q\sum 1=-3q$$
$$\sum a^4=-p\sum a^2-q\sum a=2p^2$$
$$\sum a^5=-p\sum a^3-q\sum a^2=5pq$$
Therefore $$pq=\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$$
Also,
$$\sum a^7=-p\sum a^5-q\sum a^4=-7p^2q$$
$$\implies-p^2q=\frac17\sum a^7=(pq)(-p)=\frac15\sum a^5\cdot\frac12\sum a^2$$
