# Prove that \sqrt [3]{(1+x)}-1-\frac {1}{3}x+\frac {1}{9}x^2<\frac {5}{81}x^3.

## Question:

Prove that {eq}\sqrt [3]{(1+x)}-1-\frac {1}{3}x+\frac {1}{9}x^2<\frac {5}{81}x^3 {/eq}.

## Binomial expansion application:

The binomial distribution is used to find the expanded series of a function.

The expansion of a function by binomial distribution is,

{eq}{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ..., {/eq} for n<1.

*Given

{eq}\sqrt[3]{{\left( {1 + x} \right)}} - 1 - \dfrac{1}{3}x + \dfrac{1}{9}{x^2} < \dfrac{5}{{81}}{x^3} {/eq}

Expand the expression {eq}\sqrt[3]{\left( {1 + x} \right)} {/eq} using Binomial expansion.

{eq}{\left( {1 + x} \right)^{\dfrac{1}{3}}} = 1 + \dfrac{1}{3}x + \dfrac{{\dfrac{1}{3}\left( {\dfrac{1}{3} - 1} \right)}}{2}{x^2} + \dfrac{{\dfrac{1}{3}\left( {\dfrac{1}{3} - 1} \right)\left( {\dfrac{1}{3} - 2} \right)}}{6}{x^3} + ... {/eq}

Take

{eq}f(x) = {\left( {1 + x} \right)^{\dfrac{1}{3}}} = 1 + \dfrac{1}{3}x - \dfrac{1}{9}{x^2} + \dfrac{5}{{81}}{x^3} - ... {/eq}

As f(x) has negative terms which is shown in its expansion. So, the above expression holds the below inequality.

{eq}\begin{align*} 1 + \dfrac{1}{3}x - \dfrac{1}{9}{x^2} + \dfrac{5}{{81}}{x^3} - ... & \le 1 + \dfrac{1}{3}x - \dfrac{1}{9}{x^2} + \dfrac{5}{{81}}{x^3}\\ {\left( {1 + x} \right)^{\dfrac{1}{3}}} &\le 1 + \dfrac{1}{3}x - \dfrac{1}{9}{x^2} + \dfrac{5}{{81}}{x^3}\\ {\left( {1 + x} \right)^{\dfrac{1}{3}}}-1- \dfrac{1}{3}x + \dfrac{1}{9}{x^2} & \le \dfrac{5}{{81}}{x^3} \end{align*} {/eq}

Therefore, it is proved that {eq}{\left( {1 + x} \right)^{\dfrac{1}{3}}}-1- \dfrac{1}{3}x + \dfrac{1}{9}{x^2} \le \dfrac{5}{{81}}{x^3} {/eq}.