ta có : \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow14+2\left(ab+bc+ca\right)=0\) \(\left(vìa^2+b^2+c^2=14\right)\)
\(\Rightarrow2\left(ab+bc+ca\right)=-14\)
ta có : \(\left(a^2+b^2+c^2\right)=14\Rightarrow\left(a^2+b^2+c^2\right)^2=196\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=196\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\) (1)
ta có : \(\left(2ab+2bc+2ca\right)=-14\Leftrightarrow\left(2ab+2bc+2ca\right)^2=196\)
\(\Leftrightarrow4a^2b^2+4b^2c^2+4c^2a^2+8ab^2c+8bc^2a+8ca^2b=196\)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)=196\)
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+c^2a^2\right)=196\) \(\left(vìa+b+c=0\right)\)
\(\Rightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)=\dfrac{196}{2}=98\) (2)
từ (1) và (2) ta có : \(a^4+b^4+c^2+2\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)=a^4+b^4+c^4=196-98=98\)
vậy \(a^4+b^4+c^4=98\) bởi \(a+b+c=0\) và \(a^2+b^2+c^2=14\)
