Key idea. First find the mixed products, then square the known sum of squares.

1. By C1.3,

   \[ab+bc+ca=-\frac12(a^2+b^2+c^2)=-\frac12.\]

2. Square this relation:

   \[\begin{aligned}\frac14&=(ab+bc+ca)^2\\&=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c).\end{aligned}\]

   The final term is zero, so \(a^2b^2+b^2c^2+c^2a^2=1/4\).

3. Now square \(a^2+b^2+c^2=1\):

   \[1=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2).\]

4. The mixed part equals \(2\cdot(1/4)=1/2\). Subtracting it leaves \(a^4+b^4+c^4=1/2\).

Remember. When squaring creates mixed terms, look for a second identity that evaluates those terms.