\(a^3+b^3+1=3ab\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\)
\(\Leftrightarrow\left(a+b\right)^3+1-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2+1-ab-a-b\right)=0\)
=>a+b+1=0
=>a+b=-1
\(A=\left(1+\dfrac{a}{b}\right)\left(1+b\right)\left(1+\dfrac{1}{a}\right)\)
\(=\dfrac{b+a}{b}\cdot\left(b+1\right)\cdot\dfrac{a+1}{a}\)
\(=\dfrac{-1}{b}\cdot\left(-a\right)\cdot\dfrac{-b}{a}\)
\(=-1\)
\(a^3+b^3+1=3ab\)
\(\Leftrightarrow a^3+b^3+1-3ab=0\)
\(\Leftrightarrow\left(a+b\right)^3+1-3ab-3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left[\left(a+b\right)^2-\left(a+b\right)+1\right]-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2+1-a-b-ab\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+1=0\\a^2+b^2+1-a-b-ab=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=-1\\\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a-1\right)^2+\dfrac{1}{2}\left(b-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=-1\\a=b=1\end{matrix}\right.\)
*\(a+b=-1\Rightarrow A=\left(1+\dfrac{a}{b}\right)\left(1+b\right)\left(1+\dfrac{1}{a}\right)=\dfrac{\left(a+b\right)\left(b+1\right)\left(a+1\right)}{ab}=\dfrac{-1.\left(b-a-b\right)\left(a-a-b\right)}{ab}=\dfrac{-1.\left(-a\right).\left(-b\right)}{ab}=-1\)
\(a=b=1\Rightarrow A=\left(1+\dfrac{a}{b}\right)\left(1+b\right)\left(1+\dfrac{1}{a}\right)=\left(1+\dfrac{1}{1}\right)\left(1+1\right)\left(1+\dfrac{1}{1}\right)=8\)