How prove this $x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$ show that

$$x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$$ have nice methods? Thank you 

my try:
$$\Longleftrightarrow \cos{x}\cdot x^3<(\sin{x})^3$$
let
$$f(x)=\cos{x}\cdot x^3-(\sin{x})^3$$
$$\Longrightarrow f'(x)=-\sin{x}\cdot x^3+3\cos{x}\cdot x^2-3(\sin{x})^2\cos{x}$$
 A: *

*Consider an equivalent inequality
$$
\frac{\sin x}{\sqrt[3]{\cos x}}>x.
$$

*It is enough to show that the derivative of the left-hand side is bigger than the derivative of the right hand side. After some simplification this should be (if I did not make any mistakes)
$$
2 \cos ^2 x+1-3\cos ^{4/3} x>0
$$

*Using $u=\cos^2 x$ you get 
$$
2u+1>3u^{2/3},\quad 0<u<1.
$$
The last inequality is true because $3u^{2/3}$ is increasing, concave, and the derivative at 1 is equal to 2 (make a graph).

A: We have that
$$\sin^3x-x^3 \cos x=\frac14\left(3\sin x-\sin(3x)\right)-x^3 \cos x\ge0 \iff 3\sin x-\sin(3x)-4x^3\cos x \ge 0$$
and since by Taylor's series

*

*$\sin x\ge x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7$


*$\sin (3 x)\le 3x-\frac 9 2 x^3+\frac{81}{40}x^5-\frac{243}{560}x^7+\frac{243}{4480}x^9$


*$\cos x\le 1-\frac12 x^2+\frac1{24}x^4$
then
$$ 3\sin x-\sin(3x)-4x^3\cos x\ge $$
$$\ge 3x-\frac12x^3+\frac1{40}x^5-\frac1{1680}x^7-3x+\frac 9 2 x^3-\frac{81}{40}x^5+\frac{243}{560}x^7-\frac{243}{4480}x^9-4x^3+2x^5-\frac1{6}x^8=$$
$$=-\frac1{1680}x^7+\frac{243}{560}x^7-\frac{243}{4480}x^9-\frac1{6}x^8=$$
$$=\frac{13}{30}x^7-\frac16 x^8-\frac{243}{4480}x^9\ge 0 \iff 729x^2+2240x-5824\le 0$$
which leads to $0\le x\le a\approx 1.68$.
