Ta có: $a + b + c = 0 \to \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {ab + bc + ca} \right) = 0 \leftrightarrow ab + bc + ca = \dfrac{{ - 2009}}{2}$
$ \leftrightarrow {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 2abc\left( {a + b + c} \right) = \dfrac{{{{2009}^2}}}{4} \leftrightarrow {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} = \dfrac{{{{2009}^2}}}{4}$
${\left( {{a^2} + {b^2} + {c^2}} \right)^2} = \left( {{a^4} + {b^4} + {c^4}} \right) + 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) = \left( {{a^4} + {b^4} + {c^4}} \right) + \dfrac{{{{2009}^2}}}{2} = {2009^2}$
$ \leftrightarrow \left( {{a^4} + {b^4} + {c^4}} \right) = \dfrac{{{{2009}^2}}}{2}$