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[henry0905]

Cho:$\left\{\begin{matrix} x+y+z=a & & \\ x^{2}+y^{2}+z^{2}=b^{2} & & \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c} & & \end{matrix}\right.$
Tính $x^{3}+y^{3}+z^{3}$ theo a,b,c

[thedragonknight]

Cho:$\left\{\begin{matrix} x+y+z=a(1) & & \\ x^{2}+y^{2}+z^{2}=b^{2}(2) & & \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}(3) & & \end{matrix}\right.$
Tính $x^{3}+y^{3}+z^{3}$ theo a,b,c

Ta có $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=a(b^2-\frac{xyz}{c})$(do đẳng thức (3))
Mặt khác ta có $(x+y+z)^2=a^2$
$\Leftrightarrow b^2+2xy+2yz+2xz=a^2\Rightarrow xy+yz+xz=\frac{a^2-b^2}{2}$
Ta tiếp tục có $x^2+y^2+z^2-xy-yz-xz=b^2-\frac{a^2-b^2}{2}=\frac{3b^2-a^2}{2}$
Suy ra $b^2-\frac{xyz}{c}=\frac{3b^2-a^2}{2}$
Tính đc $xyz=\frac{(a^2-b^2)c}{2}$
Vậy $a^3+b^3+c^3=\frac{3ab^2-a^3+3a^2c-3b^2c}{2}$

Edited by thedragonknight, 10-07-2012 - 19:46.