Find $f$ such that $f^{\prime}(x)=x^{2}+2$ and $f(0)=8$ \[ f(x)= \]

Step 1 ：The problem is asking for a function \(f(x)\) such that its derivative is \(x^{2}+2\) and the function evaluated at 0 is 8. To find such a function, we need to integrate the derivative function and then adjust the constant of integration so that \(f(0)=8\).

Step 2 ：Integrating the derivative function \(x^{2}+2\) gives us \(\frac{x^{3}}{3}+2x+C\), where \(C\) is the constant of integration.

Step 3 ：We then set \(f(0)=8\) to find the value of \(C\). Substituting \(0\) into the function gives us \(C=8\).

Step 4 ：Substituting \(C=8\) back into the function gives us the final function \(f(x)=\frac{x^{3}}{3}+2x+8\).

Step 5 ：\(\boxed{f(x)=\frac{x^{3}}{3}+2x+8}\) is the function that satisfies the given conditions.