The Taylor series for $f(x)=x^{3}$ at -2 is $\sum_{n=0}^{\infty} c_{n}(x+2)^{n}$. 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: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\] 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 \(n \geq 2\), 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: \[\begin{array}{l} c_{0}= \boxed{-8} \\ c_{1}= \boxed{12} \\ c_{2}= \boxed{-6} \\ c_{3}= \boxed{1} \\ c_{4}= \boxed{0} \end{array}\]