a+b+c=0
=>\(\left(a+b+c\right)^2=0\)
=>\(a^2+b^2+c^2=-2\left(ab+ac+bc\right)\)
=>\(ab+ac+bc=\frac{-\left(a^2+b^2+c^2\right)}{2}\)
a+b+c=0
=>a+b=-c
\(a^3+b^3+c^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=\left(-c\right)^3+c^3-3ab\cdot\left(-c\right)\)
=3bac
Ta có: \(\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)\)
\(=a^5+a^2b^3+a^2c^3+a^3b^2+b^5+c^3b^2+a^3c^2+b^3c^2+c^5\)
\(=\left(a^5+b^5+c^5\right)-abc\left(ab+ac+bc\right)\)
\(\Rightarrow\left(a^5+b^5+c^5\right)=\left(a^2+b^2+c^2\right)\cdot\left(a^3+b^3+c^3\right)+abc\left(ab+ac+bc\right)\)
=>\(a^5+b^5+c^5\) =\(3abc\left(a^2+b^2+c^2\right)+abc\left(ab+ac+bc\right)\)
=>\(a^5+b^5+c^5=abc\left\lbrack3\left(a^2+b^2+c^2\right)+\left(ab+ac+bc\right)\right\rbrack\)
=>\(a^5+b^5+c^5=abc\left\lbrack3\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}\right\rbrack=\frac52\cdot abc\cdot\left(a^2+b^2+c^2\right)\)
=>\(\frac52abc\left(a^2+b^2+c^2\right)=a^3+b^3+c^3=3abc\)
=>\(a^2+b^2+c^2=3:\frac52=\frac65\)
