a: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
=0
=>a^3+b^3+c^3=3abc
b: \(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2+1=0\)
=>(a-1)^2+(b+2)^2+4c^2+1=0(loại)
a.
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3+3ab.\left(-c\right)=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
b.
\(a^2-2a+b^2+4b+4c^2+6=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+4c^2+1=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+4c^2+1=0\)
Do \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\\left(b+2\right)^2\ge0\\4c^2\ge0\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2+4c^2+1>0;\forall a;b;c\)
Hay ko tồn tại a;b;c thỏa mãn đề bài
//Rất có thể đề bài ghi thiếu \(+4c\) hoặc \(-4c\)
