Answer
$1-\dfrac{x}{2}+\dfrac{3x^2}{8}$
Work Step by Step
Our aim is to write the first three terms of Maclaurin series for the given function.
We have: $f(x)=(1+x)^{-1/3}$
Consider the series for the binomial series such as: $(1+x)^k =\Sigma_{n=0}^{\infty} \dbinom{k}{n}x^n$
Now, we have:
$S=\Sigma_{n=0}^{\infty} \dbinom{-1/2}{n}x^n\\=\dbinom{-1/2}{0}x^0+\dbinom{-1/2}{1}x^1+\dbinom{-1/2}{2}x^2\\=(1)(1)+\dfrac{-1}{2} \times x-\dfrac{3}{8} x^2\\=1-\dfrac{x}{2}+\dfrac{3x^2}{8}$
