Ta có:
\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
\(=a^3+b^3+3a^2b+3ab^2\)
\(=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
Thay vào \(a^3+b^3+c^3=0\), ta được:
\(VT=a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3\)
Vì \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Leftrightarrow a^3+b^3+c^3=\left(-c\right)^{^3}-3ab\left(-c\right)+c^3\)
\(\Leftrightarrow a^3+b^3+c^3=\left(-c\right)^3+c^3+3abc\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
\(\RightarrowĐPCM\).
Có a+b+c=0
=> a+b=-c
=> (a+b)3=-c3
=> a3+b3+3ab(a+b)=-c3
=> a3+b3+3ab(-c)=-c3
=> a3+b3-3abc=-c3
=> a3+b3+c3=3abc
