\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^2-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b\right)-3\left(a+b\right).c\left(a+b+c\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b+c\right)-3\left(a+b\right).c\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3ab-3bc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Ta có:
\(a;b;c>0\)
\(\Rightarrow a+b+c>0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
\(A=2020\left(1-\dfrac{a}{b}\right)\left(1-\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)-2021\left(\dfrac{a}{b}-\dfrac{b}{c}+\dfrac{c}{a}\right)^3\)
\(\Rightarrow A=2020.\left(1-1\right)\left(1-1\right)\left(1-1\right)-2021\left(1-1+1\right)^3\)
\(\Rightarrow A=-2021\).