![](https://cdn.mathpix.com/cropped/2024_05_06_bf0205b4cbbe4b58a1adg-04.jpg?height=177&width=2027&top_left_y=526&top_left_x=10)

It is known that $a+b+c=0, a^{2}+b^{2}+c^{2}=1$. Find $a^{4}+b^{4}+c^{4}$.
$a^{4}+b^{4}+c^{4}=(a+b+c)^{4}+4 a b c(a+b+c)+2(a b+b c+a c)^{2}-4(a+b+c)^{2}(a b+b c+a c)$ (see problem $\underline{61030}$ e).

Under our conditions

$a^{4}+b^{4}+c^{4}=2 \cdot(1 / 2)^{2}=1 / 2$

## Answer

$1 / 2$.