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chắc là $c^4 $ =))
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$
$=>ab+bc+ac=\dfrac{-1}{2}$
$=>(ab+bc+ac)^2=\dfrac{1}{4}$
$=>a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=\dfrac{1}{4}$
$=>a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}$
Lại có:
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+a^2c^2)=1-2.\dfrac{1}{4}=\dfrac{1}{2}$ đpcm