1. **State the problem:** We want to find the Taylor series expansion of the function $$f(x) = x^3$$ about the point $$x=2$$ and determine the first five coefficients $$c_0, c_1, c_2, c_3, c_4$$ in the series $$\sum_{n=0}^\infty c_n (x-2)^n$$. 2. **Recall the Taylor series formula:** $$ c_n = \frac{f^{(n)}(a)}{n!} \quad \text{where } a=2 \text{ and } f^{(n)}(a) \text{ is the } n\text{-th derivative of } f \text{ evaluated at } a. $$ 3. **Calculate derivatives of $$f(x) = x^3$$:** - $$f(x) = x^3$$ - $$f'(x) = 3x^2$$ - $$f''(x) = 6x$$ - $$f^{(3)}(x) = 6$$ - $$f^{(4)}(x) = 0$$ (and all higher derivatives are zero) 4. **Evaluate derivatives at $$x=2$$:** - $$f(2) = 2^3 = 8$$ - $$f'(2) = 3 \times 2^2 = 3 \times 4 = 12$$ - $$f''(2) = 6 \times 2 = 12$$ - $$f^{(3)}(2) = 6$$ - $$f^{(4)}(2) = 0$$ 5. **Calculate coefficients using $$c_n = \frac{f^{(n)}(2)}{n!}$$:** - $$c_0 = \frac{f(2)}{0!} = \frac{8}{1} = 8$$ - $$c_1 = \frac{f'(2)}{1!} = \frac{12}{1} = 12$$ - $$c_2 = \frac{f''(2)}{2!} = \frac{12}{2} = 6$$ - $$c_3 = \frac{f^{(3)}(2)}{3!} = \frac{6}{6} = 1$$ - $$c_4 = \frac{f^{(4)}(2)}{4!} = \frac{0}{24} = 0$$ **Final answer:** $$ c_0 = 8, \quad c_1 = 12, \quad c_2 = 6, \quad c_3 = 1, \quad c_4 = 0 $$