\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=-5\)
\(\Rightarrow\left(ab+bc+ca\right)^2=25\)
\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=25\)
\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=25\)
\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]\)
\(=10^2-2.25=50\)
Ta có: a+b+c=0 ⇒(a+b+c)2=0
Hay a2+b2+c2+2ab+2bc+2ca=0
1+2(ac+bc+ca)=0
ab+bc+ca=\(\dfrac{-1}{2}\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100\left(1\right)\)
\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+b^2ac+c^2ab+a^bc=a^2b^2+b^2c^2+c^2+a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2=25\)
hay \(2\left(a^2b^2+b^2c^2+c^2a^2\right)=50\left(2\right)\)
Từ (1) và (2) ⇒a4+b4+c4=50
