\(\dfrac{a}{3}+\dfrac{b}{3}+\dfrac{c}{3}=\dfrac{a+b+c}{3}=0\)
\(\Rightarrow a+b+c=0\)
\(\dfrac{a^3}{3}+\dfrac{b^3}{3}+\dfrac{c^3}{3}=abc\)
\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{3}=\dfrac{3abc}{3}\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)\)
-\(3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)=0\)
Lại có a+b+c=0
\(\Rightarrow0\left(a^2+b^2+c^2-ac-bc-ab\right)=0\) (luôn đúng)
=>dpcm
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