Question Number 22472 by Tinkutara last updated on 18/Oct/17
$$\mathrm{If}\:{x}^{{x}} \centerdot{y}^{{y}} \centerdot{z}^{{z}} \:=\:{x}^{{y}} \centerdot{y}^{{z}} \centerdot{z}^{{x}} \:=\:{x}^{{z}} \centerdot{y}^{{x}} \centerdot{z}^{{y}} \:\mathrm{such} \\ $$$$\mathrm{that}\:{x},\:{y}\:\mathrm{and}\:{z}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{1},\:\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{x},\:{y}\:\mathrm{and}\:{z}? \\ $$$$\left(\mathrm{1}\right)\:{xyz}\:=\:\mathrm{27} \\ $$$$\left(\mathrm{2}\right)\:{xyz}\:=\:\mathrm{1728} \\ $$$$\left(\mathrm{3}\right)\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{32} \\ $$$$\left(\mathrm{4}\right)\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{12} \\ $$