Помогите решить:A^4+b^4+c^4=?Если a+b+c=0a^2+b^2+c^2=1

$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2b^2c^2+2a^2c^2+2a^2b^2=1^2\\\\ a^4+b^4+c^4=1-2(b^2c^2+a^2c^2+a^2b^2)\\\\ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=0\\\\ 2ab+2bc+2ac=-1\\\\ ab+bc+ac=-\frac{1}{2}\\\\ (ab+bc+ac)^2=\frac{1}{4}\\\\ a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=\frac{1}{4}\\ a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\\\\ a^4+b^4+c^4=1-2*\frac{1}{4}=\frac{1}{2}$

Ответ $\frac{1}{2}$