Brilliant problem! 1) Denote $a={{\log }_{2}}3,\text{ }b={{\log }_{3}}5,\text{ }c={{\log }_{5}}2$ Using the change of the basis of the logarithm, we have: 2) $abc=1$ and 3) $ab+bc+ca={{\log }_{2}}5+{{\log }_{3}}2+{{\log }_{5}}3$ 4) So our equation is: ${{x}^{3}}-(a+b+c){{x}^{2}}+(ab+bc+ca)x-abc=0\Leftrightarrow (x-a)(x-b)(x-c)=0$ 5) So the solutions there are: ${{x}_{1}}={{\log }_{2}}3,\text{ }{{x}_{2}}={{\log }_{3}}5,\text{ }{{x}_{3}}={{\log }_{5}}2$. Done

Can't we explict it like $a=log3,b=log2,c=log5$ And use the identity $\frac{loga}{logb}$ for log a base b to get solution. By vista the problem is easy just check the cases $p+q+r=t$ ,$t=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ And $pqr=1$,$pq+qr+rp=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ it implies the solution Just realized the above solution is similar.

Yes, the this solution it exactly the same that my solution!

Just Viete's formulas.

Yes, just Viete formulas and other 99% idea!