2017/033) If $a+b+c+d+e+f=0$ and $a^3+b^3+c^3+d^3+e^3+f^3=0$ and no 2 variables are additive inverse of each other then show that $(a+c)(a+d)(a+e)(a+f) = (b+c)(b+d)(b+e)(b+f)$

We have $(a+c+d) = - (b+e+f)\cdots(1)$
and $a^3+c^3+d^3= -(b^3+e^3+f^3)\cdots(2)$
cube both sides of (1) using $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$  to get
$a^3+c^3+ d^3 + 3(a+c)(a+d)(d+c) = -(b^3 + e^3 + f^3 + 3(b+e)(b+f)(e+f)$
or $(a+c)(a+d)(c+d) = - (b+e)(b+f)(e+f)\cdots(3)$ using (2)
similarly we have $(a+e)(a+f)(e+f) = - (b+c)(b+d)(c+d)\cdots(4)$
multiplying (3) and (4) we get the result