# If a^3+ b^3 + c^3 = 3abc find the value of a + b + c

$\\ Given, \ a^{3}+b^{3}+c^{3}=3 a b c \\ We \ know\\ a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\\ Now\ a^{3}+b^{3}+c^{3}=3 a b c,\\ So\ 3 a b c-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\\ Or, (a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0 \\ \mathrm{So} \ a^{2}+b^{2}+c^{2}-a b-b c-c a=0 \ Or \ a+b+c=0$

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