The Taylor series for $f(x)=x^3$ at -2 is $\sum _{n=0}^{\mathrm{\infty }}{c}_n(x+2{)}^n$. Find the first few coefficients.
$$\begin{array}{l}{c}_0=\\ c_1=\\ c_2=\\ c_3=\\ c_4=\end{array}$$

Step 1 ：The Taylor series for a function $f(x)$ about a point $a$ is given by: $$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$ where $f^{(n)}(a)$ is the nth derivative of $f$ evaluated at $a$.

Step 2 ：In this case, $f(x)=x^3$, $a=-2$, and we need to find the coefficients $c_n$ for $n=0,1,2,3,4$.

Step 3 ：The nth derivative of $f(x)=x^3$ is $f^{(n)}(x)=3n(n-1)x^{n-2}$ for $n\ge 2$, and $f^{(n)}(x)=x^3$ for $n=0$, $f^{(n)}(x)=3x^2$ for $n=1$.

Step 4 ：We can substitute these into the formula for the Taylor series to find the coefficients.

Step 5 ：The coefficients of the Taylor series for $f(x)=x^3$ at $a=-2$ are $c_0=-8$, $c_1=12$, $c_2=-6$, $c_3=1$, and $c_4=0$. These are the values of the function and its derivatives at $x=-2$, divided by the factorial of the order of the derivative.

Step 6 ：Final Answer: $$\begin{array}{l}{c}_0=\boxed{{\displaystyle -8}}\\ c_1=\boxed{{\displaystyle 12}}\\ c_2=\boxed{{\displaystyle -6}}\\ c_3=\boxed{{\displaystyle 1}}\\ c_4=\boxed{{\displaystyle 0}}\end{array}$$