Find the first four terms of the Maclaurin series for 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
1+x3−x29+5x3811 + \frac{x}{3} - \frac{x^2}{9} + \frac{5x^3}{81}1+3x−9x2+815x3
1+x3−x218+x3811 + \frac{x}{3} - \frac{x^2}{18} + \frac{x^3}{81}1+3x−18x2+81x3
1+x3+x29−5x3811 + \frac{x}{3} + \frac{x^2}{9} - \frac{5x^3}{81}1+3x+9x2−815x3
1+x3−x26+x3271 + \frac{x}{3} - \frac{x^2}{6} + \frac{x^3}{27}1+3x−6x2+27x3