(a+b+c)^3=a^3+b^3+c^3+3*(a+b+c)*(ab+bc+ac)-3abc
x^3 + y^3+ z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) + 3xyz
(a+b)(b+c)(c+a)=(a+b+c)*(ab+bc+ac)-abc
(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf
(x-1)(y-1)(z-1)=xyz-xy-xz-yz+x+y+z-1
(x+1)(y+1)(z+1)=xyz+xy+xz+yz+x+y+z+1
Este problema no podia faltar:
http://www.qbyte.org/puzzles/p079s.html
http://mathworld.wolfram.com/Newton-GirardFormulas.html
http://planetmath.org/encyclopedia/ProofOfNewtonGirardFormulaForSymmetricPolynomials.html
http://en.wikipedia.org/wiki/Newton_identities#Computing_power_sums
http://www.numericana.com/answer/algebra.htm#sym
http://www.mathpages.com/home/kmath097.htm
http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums
En la página 8 esta la recurrencia general:
http://www.artofproblemsolving.com/Resources/Papers/PolynomialsAK.pdf
http://mathworld.wolfram.com/VietasFormulas.html
http://www.obm.org.br/opencms/revista_eureka/ (Revista nº 25)
León-Sotelo

2 comentarios:
((n^2+n)/2)^2-((n^2-n)/2)^2=n^3
_:) GRACIAS!
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