Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.

$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant
 \sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1 $

Sadly $\ 6\sqrt[6]{\frac{1}{108}}-1$ is not correct answer, it is not the minimum.

Very similar to what you have done:
$$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}}{5}\geq
 \sqrt[5]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[5]{\frac{1}{108}}$$
This gives us
$$x^3+\frac{1}{x^2}\geq \sqrt[5]{\frac{1}{108}}$$
Desmos screenshot:

P.S. Your method fails because equality holds only when
$$\frac{x^3}{2}=\frac{1}{3x^2}=1$$ which is impossible. The extra "one" you added in your AM-GM application screwed your attempt.
