Từ \(a+b+c=0\Rightarrow\left(a+b+c\right)^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow2ab+2bc+2ca=-14\Rightarrow ab+bc+ca=-7\)
Hay \(\left(ab+bc+ca\right)^2=49\)\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=49\). Lại có:
\(A=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(=14^2-2\cdot49=196-98=98\)