Use the binomial series (1+x)r=∑n=0∞(rn)xn(1+x)^r = \sum_{n=0}^{\infty} \binom{r}{n} x^n(1+x)r=∑n=0∞(nr)xn to expand f(x)=1+x3=(1+x)1/3f(x) = \sqrt[3]{1+x} = (1+x)^{1/3}f(x)=31+x=(1+x)1/3. What is the coefficient of x2x^2x2?
(1/32)=(1/3)(−2/3)2!=−19\binom{1/3}{2} = \frac{(1/3)(-2/3)}{2!} = -\frac{1}{9}(21/3)=2!(1/3)(−2/3)=−91
19\frac{1}{9}91
127\frac{1}{27}271
−127-\frac{1}{27}−271