problem to make you happy (2)

We note that:

$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)$

$\Rightarrow a^4+b^4+c^4 = (a^2+b^2+c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2)$

$\quad \quad \quad \quad \quad \quad \quad = 1 - 2(a^2b^2 + b^2c^2 + c^2a^2)$

Now, we need to find: $\quad a^2b^2 + b^2c^2 + c^2a^2$

We also note that:

$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab + bc + ca)$

$\Rightarrow ab+bc+ca = \dfrac {(a+b+c)^2 - (a^2+b^2+c^2)} {2} = \dfrac {0 - 1} {2} = -\frac {1}{2}$

And that:

$(ab+bc+ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2(ab^2c + abc^2 + a^2bc)$

$\quad \quad \quad \quad \quad \quad \quad = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c)$

$\quad \quad \quad \quad \quad \quad \quad = a^2b^2 + b^2c^2 + c^2a^2 + 0)$

$\Rightarrow a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 = \left( -\frac {1}{2} \right)^2 = \frac {1}{4}$

$\Rightarrow a^4+b^4+c^4 = 1 - 2(a^2b^2 + b^2c^2 + c^2a^2)$

$\quad \quad \quad \quad \quad \quad \quad = 1 - 2\left( \frac {1}{4} \right) = 1 - \frac {1}{2} = \boxed {\frac {1}{2}}$