Answer
$1-\dfrac{4}{3}x+\dfrac{2}{3}x^2-\dfrac{4}{27}x^3$
Work Step by Step
Apply Binomial series formula to find the first four terms.
$(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$
and $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$
Thus, we have
$(1 -\dfrac{x}{3})^{4}=1+(4)(-\dfrac{x}{3})+\dfrac{(4)(3)(\dfrac{-x}{3})^2}{2!}+\dfrac{(4)(3)(2)(\dfrac{-x}{3})^3}{3!}+...$
or, $=1-\dfrac{4}{3}x+\dfrac{(4)(3)(\dfrac{1}{9})x^2}{2!}+\dfrac{(4)(3)(2)(\dfrac{-1}{27})x^3}{3!}+...=1-\dfrac{4}{3}x+\dfrac{2}{3}x^2-\dfrac{4}{27}x^3+..$
Thus, the first four terms are: $1-\dfrac{4}{3}x+\dfrac{2}{3}x^2-\dfrac{4}{27}x^3$
