Question Number 176448 by peter frank last updated on 19/Sep/22
$$\mathrm{Suppose}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \\ $$$$=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{abc}=\mathrm{0} \\ $$$$ \\ $$
Commented by mr W last updated on 19/Sep/22
$${you}\:{can}'{t}\:{prove}!\:{because}\:{it}'{s}\:{not}\:{true}. \\ $$$${example}\:{a}={b}={c}=\mathrm{1}. \\ $$
Commented by mr W last updated on 19/Sep/22
$${suppose}\: \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}+{b}+{c}={k} \\ $$$${we}\:{can}\:{get} \\ $$$${abc}=\frac{{k}\left({k}−\mathrm{1}\right)\left({k}−\mathrm{2}\right)}{\mathrm{6}}. \\ $$
Commented by peter frank last updated on 19/Sep/22
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{clarification} \\ $$