Cube Root Inequality

$\begin{aligned} 4^\frac 13 & = 2^\frac 23 \\ & = (1+1)^\frac 23 & \small \color{#3D99F6} \text{By binomial expansion} \\ & = 1 + \frac 23 - \frac {\frac 23 \cdot \frac 13 }{2!} + \frac {\frac 23 \cdot \frac 13 \cdot \frac 43}{3!} - \cdots \\ & = {\color{#3D99F6}1 + \frac 23 - \frac 19} + \frac 4{81} - \cdots \end{aligned}$

$\begin{aligned} \implies 4^\frac 13 > {\color{#3D99F6}1 + \frac 23 - \frac 19} = \frac {14}9 \approx 1.555 > 1.5 \end{aligned}$

If $a^3>b^3$ , then $a>b$ , where a, b are real numbers.

From this let $a = 4^{1/3}$ and $b = 1.5$

$a^3 = 4$ and $b^3 =3.375$

Clearly $a^3>b^3$ which means $a>b$ , so $4^{1/3}>1.5$

$Suppose\ that\ 4^{\frac {1}{3}}>1.5$

$So\ 4^{\frac {1}{3}}>{\frac {3}{2}}$

$By\ cubing\ both\ sides\ we\ conclude\ that\ 4>{\frac {27}{8}}$

$So\ 32>27$

$**From\ that\ we\ conclude\ that\ the hypothesis\ is\ true.**$