ta có: \(x+y+z=a\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=a^2\)

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

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{c}\Rightarrow c\left(xy+yz+xz\right)=xyz\)

Ta có:\(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)

\(=a\left(b-\frac{a^2-b}{2}\right)+\frac{3c\left(a^2-b\right)}{2}\)

Ta có:

\(x+y+z=a\)

\(\Rightarrow\left(x+y+z\right)^2=a^2\)

Ta lại có:

\(x^2+y^2+z^2=b^2\)

\(\Rightarrow\left(x+y+z\right)^2-\left(x^2+y^2+z^2\right)=a^2-b^2\)

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

\(\Rightarrow2\left(xy+xz+yz\right)=a^2-b^2\)

\(\Rightarrow xy+xz+yz=\dfrac{a^2-b^2}{2}\left(1\right)\)

Lại có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=c\)

\(\Rightarrow\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=c\)

\(\Rightarrow\dfrac{yz+xz+xy}{xyz}=c\)

\(\Rightarrow yz+xz+xy=c.xyz\left(2\right)\)

Từ (1) và (2) suy ra:

\(\dfrac{a^2-b^2}{2}=c.xyz\)

\(\Rightarrow\dfrac{a^2-b^2}{2c}=xyz\)

Như vậy ta có:

\(\left\{{}\begin{matrix}x+y+z=a\\xy+yz+zx=\dfrac{a^2-b^2}{2}\\xyz=\dfrac{a^2-b^2}{2c}\end{matrix}\right.\)

Ta có:

\(x^3+y^3+z^3\)

\(=\left(x+y+z\right)^3-3\left(x^2z+xyz+xz^2+x^2y+xyz+xy^2+y^2z+xyz+yz^2\right)+3xyz\)

\(=\left(x+y+z\right)^3-3\left[xz\left(x+y+z\right)+xy\left(x+y+z\right)+yz\left(x+y+z\right)\right]+3xyz\)

\(=\left(x+y+z\right)^3-3\left[\left(xy+yz+zx\right)\left(x+y+z\right)\right]+3xyz\)

\(=a^3-3\left[\dfrac{\left(a^2-b^2\right)}{c}.a\right]+3\left(\dfrac{a^2-b^2}{2c}\right)\)

\(=a^3-\dfrac{3a\left(a^2-b^2\right)}{c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)

\(=a^3-\dfrac{6a\left(a^2-b^2\right)}{2c}+\dfrac{3\left(a^2-b^2\right)}{2c}\)

\(=a^3-\dfrac{6a\left(a^2-b^2\right)+3\left(a^2-b^2\right)}{2c}\)

\(=a^3-\dfrac{3\left(a^2-b^2\right)\left(2a+1\right)}{2c}\)

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