Originally Posted by

**mark1950** I have a problem with this particular question:

Expand $\displaystyle (\frac{1 + x}{1 - 3x})^{\frac{1}{3}} $ in ascending powers of x up to and including the term in $\displaystyle x^3 $ and determine the set of values of x for this expansion to be valid. Calculate the value of $\displaystyle \sqrt[3] 3 $ giving your answer correct to 3 decimal places.

This is my take on the question:

$\displaystyle =(1 + x)^{\frac{1}{3}}(1 - 3x)^{-\frac{1}{3}} $

$\displaystyle =[ 1 + \frac{1}{3} + \frac{(\frac{1}{3})(\frac{1}{3} - 1)}{2!}x^2 + \frac{(\frac{1}{3})(\frac{1}{3} - 1)(\frac{1}{3} - 2)}{3!}x^3 + ...]$$\displaystyle [1 + (-\frac{1}{3})(-3x) + \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)}{2!} + (-\frac{1}{3})(-3x)

+ \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)(-\frac{1}{3} - 2)}{3!} + ...] $

$\displaystyle =[ 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ... ][1 + x + 2x^2 + \frac{14}{3}x^3] $

$\displaystyle =[1 + x + 2x^2 + \frac{14}{3}x^3 + \frac{1}{3}x + \frac{1}{3}x^2 + \frac{2}{3}x^3 - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ...] $

$\displaystyle =[1 + \frac{4}{3}x + \frac{20}{9}x^2 + \frac{428}{81}x^3]$

I got the above expression but when I try to substitute $\displaystyle x = \frac{1}{5}$ into the expression to solve the last part of the question, the value aren't the same and the difference is quite large (mine was 1.39) compared with the standard value of 1.442 (value of $\displaystyle \sqrt[3] 3$). Where have I gone wrong? Thanks.