ta có:
a+b+c=0 \Rightarrow $(a+b+c)^2$ =0
\Leftrightarrow $a^2$ + $b^2$ + $c^2$ + 2(ab+bc+ca) =0
\Leftrightarrow 14 + 2(ab+bc+ca) = 0
\Leftrightarrow 2(ab+bc+ca) = -14
\Rightarrow ab+bc+ca = -7
$a^2$ +$b^2$ + $c^2$ = 14
\Rightarrow ($a^2$ +$b^2$ + $c^2$)^2 = 196
\Leftrightarrow $a^4$ +$b^4$ + $c^4$ + 2($a^2b^2$+$b^2c^2$+$c^2a^2$) = 196
\Leftrightarrow$a^4$ +$b^4$ + $c^4$ +2[$(ab+bc+ca)^2$ - 2(a+b+c)] = 196
\Leftrightarrow$a^4$ +$b^4$ + $c^4$ + 2.(-7)^2 = 196
\Leftrightarrow $a^4$ +$b^4$ + $c^4$ = 196 - 98 = 98
vậy, $a^4$ +$b^4$ + $c^4$ = 98