Hi Guys,

Need a little hint on this one.

Expand $\displaystyle (1 + x)^{\frac{1}{3}}$ in ascending powers of $\displaystyle x$ as far as the term in $\displaystyle x^3$, simplifying the terms as much as possible. By substituting 0.08 for $\displaystyle x$ in your result obtain an approximate value of the cube root of 5, upto four decimal places.

I expanded the problem as a binomial series,

$\displaystyle 1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \dfrac{5x^3}{81}$

The rest of the question is confusing me. How do I break up 5 to include 0.08?

The only approach I could think of was,

$\displaystyle

\sqrt[3]{5} = \sqrt[3]{8\left(1 - \frac{3}{8}\right)} = 2\sqrt[3]{1 - \frac{3}{8}}

$

Then using x = -3/8, to put in the above equation.

This gives me 1.7122 which is incorrect, and also does not use the suggested 0.08. Any ideas where I am going wrong?

Thanks for your help.