Question

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

Answer 1

\(a=x+y+z\Rightarrow a^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=b+2\left(xy+yz+xz\right)\Rightarrow xy+yz+xz=\frac{a^2-b}{2}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow3c.\frac{a^2-b}{2}=3xyz\)

Ta có:

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=a\left(b-\frac{a^2-b}{2}\right)-3c.\frac{a^2-b}{2}\)