Ta có:
\(xy+yz+zx=-5;xz=-5\)
\(\Rightarrow xy+yz=0\)
\(\Rightarrow y\left(x+z\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=0\\x+z=0\end{cases}}\)
Nếu \(y=0\) ta có:
\(x+0+z=2\Rightarrow x+z=2\)
\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=2\cdot\left(2^2+3\cdot5\right)+0=38\)
Nếu \(x+z=0\Rightarrow y=2\),ta có:
\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=8\)
Vậy \(A=8\left(h\right)A=38\)
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