Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(\frac{1}{a}+\frac{1}{b}\right)=0\)
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(\frac{1}{-c}\right)=0\)
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=3\)
\(\Rightarrow A=3\)
Vậy \(A=3\)
Tham khảo nhé~
