\(a-b=1\Leftrightarrow a=b+1\)
\(A=\left(b+1\right)^3-b^3-b\left(b+1\right)\)
\(A=b^3+3b^2+3b+1-b^3-b^2-b\)
\(A=2b^2+2b+1=\frac{1}{2}\left(4b^2+4b+1\right)+\frac{1}{2}=\frac{1}{2}\left(2b+1\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
\(A_{min}=\frac{1}{2}\) khi \(\left\{{}\begin{matrix}a=\frac{1}{2}\\b=-\frac{1}{2}\end{matrix}\right.\)