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Assume we have 6 real numbers $a,b,c,d,e,f$ such that:$$a+b+c+d+e+f=0$$and $$a^3+b^3+c^3+d^3+e^3+f^3=0$$prove that $$(a+c)(a+d)(a+e)(a+f)=(b+c)(b+d)(b+e)(b+f)$$Is there any way to prove this without direct multiplication? i.e. is there any cute way to attain that?

Thanks in advance!

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  • $\begingroup$ Using only $a+b+c+d+e+f=0$ I have reached the equality $$(a+d)(b+d)(c+d)=(e+f)(de+df+ef-bc-2ab)$$ Don't know how to continue though. $\endgroup$
    – TheSimpliFire
    Jan 30, 2018 at 17:39

1 Answer 1

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From conditions we obtain: $$a+b=-c-d-e-f$$ and $$ab=\frac{\sum\limits_{cyc}(c^2(d+e+f)+2cde)}{c+d+e+f}$$ and we need to prove that $$a^4+a^3\sum_{cyc}c+\frac{1}{4}a^2\sum_{sym}cd+a\sum_{cyc}cde+cdef=$$ $$=b^4+b^3\sum_{cyc}c+\frac{1}{4}b^2\sum_{sym}cd+b\sum_{cyc}cde+cdef,$$ for which we need to prove that

$$(a+b)(a^2+b^2)+(a^2+ab+b^2)\sum_{cyc}c+\frac{1}{4}(a+b)\sum_{sym}cd+\sum_{cyc}cde=0$$ or $$(a+b)(a^2+b^2)-(a^2+ab+b^2)(a+b)-\frac{1}{4}(c+d+e+f)\sum_{sym}cd+\sum_{cyc}cde=0$$ or $$-ab(a+b)-\frac{1}{4}(c+d+e+f)\sum_{sym}cd+\sum_{cyc}cde=0$$ or $$\sum\limits_{cyc}(c^2(d+e+f)+2cde)-\frac{1}{4}(c+d+e+f)\sum_{sym}cd+\sum_{cyc}cde=0,$$ which is obvious.

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