Given $a+b+..=a^7+b^7+..=0$ show that $a(a+b)..=0$ Question: 
Suppose $a,b,c,d$ are real numbers such that
$a+b+c+d=a^7+b^7+c^7+d^7=0$
Show that
$a(a+b)(a+c)(a+d)=0$
My attempt: Using $a+b+c+d=0$, I get
$a(a+b)(a+c)(a+d)= 0
\implies a=0, \text{or}$ $ a(bc+cd+db)+bcd=0$
How can I use $a^7+b^7+c^7+d^7=0$ to prove $a(bc+cd+db)+bcd=0$ ?
Edit:(courtesy @mathguy) 
Replacing $d$ by $-(a+b+c)$ we see that the hypothesis is equivalent to 
$(a+b+c)^7=a^7+b^7+c^7$, and the conclusion equivalent to $a(a+b)(a+c)(b+c)=0$. The polynomial $(a+b+c)^7-a^7-b^7-c^7$ is divisible by $(a+b)(a+c)(b+c)$, and another irreducible fourth degree symmetric polynomial $P(a,b,c)$ .
Also, $P(0,b,c)= b^4+2b^3c+3b^2c^2+2bc^3+c^4$. It remains to be shown that $P(a,b,c)=0$ when $a=0$
 A: Since $d=-(a+b+c)$ and $d^7=-(a^7+b^7+c^7)$, we obtain
$$(a+b+c)^7-a^7-b^7-c^7=0$$ or
$$7(a+b)(a+c)(b+c)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$$ or
$$(a+b)(a+c)(a+d)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0.$$
Thus, it remains to prove that if $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=0$.
We'll prove that $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)\geq0$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$,where $v^2$ can be negative, and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for en extremal value of $w^3$.
Since $a$, $b$ and $c$ are real roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x=w^3$, 
we see that the graph of $g(x)=x^3-3ux^2+3v^2x$ and a line $y=w^3$ have three common points
and $w^3$ gets an extremal value, when a line $y=w^3$ is a tangent line to the graph of $g$,
which happens for equality case of two variables.
Since our inequality is symmetric, we can assume $c=b$,
which gives $a^4+4a^3b+11a^2b^2+14ab^3+9b^4\geq0$ or
$$(a+b)^4+ 5(a+b)^2b^2+3b^4\geq0,$$
which is obvious.
The equality occurs for $a=b=0$ and we proved that 
if 
$\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=b=c=0$ 
and we are done!
