Question

If $a+b+c=0,a^2+b^2+c^2=4$, then $a^4+b^4+c^4$ is _____

Answer 1

Answer: 8

$(a+b+c{)}^2=0$
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)=0$
$\Rightarrow ab+bc+ca=-2$
Squaring
$a^2{b}^2+b^2{c}^2+c^2{a}^2+2a{b}^{2}c+2a^{2}bc+2ba{c}^2=4$
$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc(a+b+c)=4$
$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=4$
Now $a^2+b^2+c^2=4$
Squaring, we get
$a^4+b^4+c^4+2\left(a^2{b}^2+a^2{c}^2+b^2{c}^2\right)=16$
or $a^4+b^4+c^4=8$