Question

xyz=?
x^{3}+ y^{3}+z^{3}= x^{2}+y^{2}+z^{2}=x+y+z[/tex]

Answer 1

(x+y+z)²=x²+y²+z²+2(xy+yz+zx) Cum x²+y²+z²=x+y+z⇒
(x+y+z)²=(x+y+z)+2(xy+yz+zx) Fie asta ecuatia (1)
(x+y+z)³=(x+y+z)²·(x+y+z)
            =[(x+y+z)+2(xy+yz+zx)]·(x+y+z)
            = (x+y+z)² +2(xy+yz+zx)·(x+y+z)
(x+y+z)³ =(x+y+z)+2(xy+yz+zx)+2(xy+yz+zx)·(x+y+z) Fie asta ecuatia (2)

Se descompune facand inmultire termen cu termen (x+y+z)³.
(x+y+z)³=x³+y³+z³+3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz Cum x³+y³+z³=x+y+z
(x+y+z)³=(x+y+z)+3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz Fie asta ecuatia (3)

Din ecuatiile 2 si 3 egalezi termenii din dreapta.
(x+y+z)+3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz=(x+y+z)+2(xy+yz+zx)+2(xy+yz+zx)·(x+y+z)
3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz=2(xy+yz+zx)+2(xy+yz+zx)·(x+y+z)
3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz=2(xy+yz+zx)+2(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz

x²y+x²z+y²x+y²z+z²x+z²y=2(xy+yz+zx) Fie asta ecuatia (4)
Pentru simplitatea scrierii notez cu A=x+y+z
Atunci ecuatia 3 devine:
A³=A+3(x²y+x²z+y²x+y²z+z²x+z²y)+6xyz
Din ecuatia 4⇒x²y+x²z+y²x+y²z+z²x+z²y=2(xy+yz+zx)
A³=A+3·2(xy+yz+zx)+6xyz
Din ecuatia 1⇒2(xy+yz+zx)=A²-A
A³=A+3·(A²-A)+6xyz

A³-3A²+2A=6xyz
xyz =A(A-1)(A-2)/6 unde A=x+y+z