Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
=>\(\frac{1}{z}=-\left(\frac{1}{x}+\frac{1}{y}\right)\)
=>\(\left(\frac{1}{z}\right)^3=\left[-\left(\frac{1}{x}+\frac{1}{y}\right)\right]^3\)
=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}+\frac{1}{y}\right)^3\right]\)
=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}\right)^3+3.\left(\frac{1}{x}\right)^2.\frac{1}{y}+3.\frac{1}{x}.\left(\frac{1}{y}\right)^2+\left(\frac{1}{y}\right)^3\right]\)
=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{x}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{y}+\frac{1}{y^3}\right]\)
=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{y^3}\right]\)
=>\(\frac{1}{z^3}=-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left\{-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\right\}\)
\(\frac{1}{z^3}-\left[-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)\right]=-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\)
Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=-3.\frac{1}{x}.\frac{1}{y}.\left(-\frac{1}{z}\right)\)
=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=3.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}\)
=>\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=3.\frac{1}{xyz}\)
=>\(xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=3\)
=>A=3
Vậy A=3
