Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$

Let $T_{m}$ be $a^m+b^m+c^m$.

Let $k=-ab-bc-ca$, and $l=abc$.

Note that this implies $a,b,c$ are solutions to $x^3=kx+l$.

Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)

It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$.

From here, note that $T_{4}=2k^2$ using the identity above.

In the same method, note that $T_{5}=5kl$.

From here, note $T_{7}=5k^2l+2k^2l=7k^2l$ from $T_{m+3}=kT_{m+1}+lT_{m}$ . Therefore, the equation simplifies to showing that $k^2l \times l=(kl)^2$, which is true.

Useful identities:

$(y-z)^{3}+(z-x)^{3}+(x-y)^{3}= 3(y-z)(z-x)(x-y)$

$(y-z)^{5}+(z-x)^{5}+(x-y)^{5}= 5(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)$

$(y-z)^{7}+(z-x)^{7}+(x-y)^{7}= 7(y-z)(z-x)(x-y)(x^{2}+y^{2}+z^{2}-yz-zx-xy)^{2}$

By letting $a=y-z,b=z-x,c=x-y$.

Start off with the fact that $$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = - 2(ab+bc+ca) $$

Since $a+b+c=0$ we know that ${a,b,c}$ are the three roots of some cubic missing the $x^2$ term: $$ x^3+kx+m = 0 $$ with $ab+bc+ca = k$. And by the way, that says that $$a^2+b^2+c^2=-2k$$

Now start chaining upward, expressing $a^n+b^n+c^n$ in terms of $m$ and $k$. For $n=3$ we add the cubic expression with $x=a$ to that with $x=b$ and $x=c$ to get $$ a^3+ b^3 + c^3 +k(a+b+c)+(m+m+m) = 0 \\ a^3+ b^3 + c^3 +3m = 0 \\ a^3+ b^3 + c^3 = -3m $$ Now for $n=4$ we multiply each of the equations by $a$, $b$ and $c$ respectively before adding them: $$ a^4+ b^4 + c^4 +k(a^2+b^2+c^2)+m(a+b+c) = 0 \\ a^4+ b^4 + c^4 -2k^2 = 0 \\ a^4+ b^4 + c^4 = 2k^2 $$ Now for $n=5$ we multiply each of the equations by $a^2$, $b^2$ and $c^2$ respectively before adding them: $$ a^5+ b^5 + c^5 +k(a^3+b^3+c^3)+m(a^2+b^2+c^2) = 0 \\ a^5+ b^5 + c^5 -3mk -2mk = 0 \\ a^5+ b^5 + c^5 = 5mk $$ For this problem we can afford to skip $n=6$. $$ a^7+ b^7 + c^7 +k(a^5+b^5+c^5)+m(a^4+b^4+c^4) = 0 \\ a^7+ b^7 + c^7 +5mk^2 +2mk^2 = 0 \\ a^7+ b^7 + c^7 = -7mk^2 $$ Then your identity reads $$ \left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\ \left( -m \right) \left( -mk^2 \right) = (mk)^2 $$