\(\frac{n^3-1}{n^3+1}=\frac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow A=\frac{1\left(3^2-3+1\right)}{3\left(2^2-2+1\right)}.\frac{2.\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}.\frac{3\left(5^2-5+1\right)}{5.\left(4^2-4+1\right)}...\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(=\frac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right).n\left(n+1\right)}=\frac{2\left(n^2+n+1\right)}{3\left(n^2+n\right)}\)
