Key idea. Use the zero sum twice: first by squaring, then by substituting for c.

1. Square \(a+b+c=0\):

   \[a^2+b^2+c^2+2ab+2bc+2ca=0.\]

   Rearranging proves \(a^2+b^2+c^2=-2(ab+bc+ca)\).

2. For the cubes, write \(c=-(a+b)\). Then

   \[a^3+b^3+c^3=a^3+b^3-(a+b)^3.\]

3. Expand the last cube and cancel:

   \[a^3+b^3+c^3=-3a^2b-3ab^2=-3ab(a+b).\]

4. Since \(a+b=-c\), this equals \(3abc\), proving the second identity.

Remember. A relation between variables can make otherwise complicated expansions collapse.