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