Question

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

Answer 1

$( 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}+2 a b^{2} c+2 a^{2} b c+2 b a c^{2}=4 $
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=4$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=4$
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 $a^{4}+b^{4}+c^{4}=8$