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