A) \[{{a}^{2}}c+{{b}^{3}}+a{{c}^{2}}=3abc\]
B) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]
C) \[{{b}^{2}}c+{{c}^{2}}a+{{a}^{2}}b=abc\]
D) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca=0\]
Correct Answer: A
Solution :
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