\(\frac{d}{{dx}}{({x^3} + 1)^9} = \) \(9{({x^3} + 1)^8}(3{x^2})\) \(9{({x^3} + 1)^8}\) \(9{x^2}{(3{x^2} + 1)^8}\) \(\frac{{({x^3} + 1){}^9}}{{3{x^2}}}\)