${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

$\Leftrightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0$

$\Leftrightarrow {{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-(3{{a}^{2}}b+3a{{b}^{2}})+{{c}^{3}}-3abc=0$

$\Leftrightarrow {{(a+b)}^{3}}+{{c}^{3}}-3ab(a+b+c)=0$

$\Leftrightarrow (a+b+c)\left[ {{(a+b)}^{2}}-(a+b).c+{{c}^{2}} \right]-3ab(a+b+c)=0$

$\Leftrightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-ac-bc)-3ab(a+b+c)=0$

$\Leftrightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-ac-bc-3ab)=0$

$\Leftrightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ac-bc-ab)=0$

$\Leftrightarrow (a+b+c)(2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ac-2bc-2ab)=0$

$\Leftrightarrow (a+b+c)\left[ {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-b)}^{2}} \right]=0$

$\Leftrightarrow \left[ \begin{align}& a+b+c=0 \\ & {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-b)}^{2}}=0 \\ \end{align} \right.$

$\Leftrightarrow \left[ \begin{align}& a+b+c=0 \\ & a=b=c \\ \end{align} \right.$