Tricky exercise asked in the mathematics teacher group

Exercise:

Solve in $\mathbb{R}$ , ${{x}^{2{{x}^{2}}-3x+1}}={{x}^{3}}$

Solution: we have ${{a}^{b}}={{a}^{c}}\,\,iff\,\,b=c\,\,$

$\left( 2{{x}^{2}}-3x+1 \right)\ln x-3\ln x=0\Leftrightarrow \ln x\left( 2{{x}^{2}}-3x-2 \right)=0\Leftrightarrow \ln x=0\,\,or\,\,2{{x}^{2}}-3x-2=0$

$\ln x=\ln 1\Leftrightarrow x=1\,\,\,or\,\,\,2{{x}^{2}}-3x-2=0$

So $2{{x}^{2}}-3x+1=3\Leftrightarrow 2{{x}^{2}}-3x+1-3=0\Leftrightarrow 2{{x}^{2}}-3x-2=0$

$\Leftrightarrow {{x}^{2}}-\frac{3}{2}x-1=0\Leftrightarrow {{x}^{2}}-2\left( \frac{3}{4} \right)x-1=0\Leftrightarrow {{x}^{2}}-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16}-1=0$

$\Leftrightarrow {{\left( x-\frac{3}{4} \right)}^{2}}=\pm \frac{5}{4}\Leftrightarrow x=\frac{3\pm 5}{4}\Leftrightarrow x=2\,or\,x=\frac{-1}{2}$

${{x}^{3}}-{{x}^{2{{x}^{2}}-3x+1}}=0\Leftrightarrow x\left( {{x}^{2}}-{{x}^{2{{x}^{2}}-3x}} \right)=0\Leftrightarrow x=0\,\,or\,\,{{x}^{2}}-{{x}^{2{{x}^{2}}-3x}}=0$

The real roots are $\left\{ \frac{-1}{2},0,1,2 \right\}$