Choi Lam homogeneous polynomials as sums of squares
Question 1 can be done by hand pretty simply.
$$ S_8 = (x^2 + y^2 + z^2) (x^4 y^2 + y^4 z^2 + z^4 x^2 - 3 x^2 y^3 z^2) $$
So
$$ = x^6 y^2 + x^4 y^4 - 2 x^4 y^2 z^2 + \text{cyclic cycle indices} $$
Observe that
$$ x^6 y^2 - 2 x^4 y^2 z^2 + x^2 y^2 z^4 = x^2 y^2 (x^2 - z^2)^2 $$
And that
$$ \frac12 z^4 y^4 + \frac12 x^4 z^4 - x^2 y^2 z^4 = \frac12 z^4 (x^2 - y^2)^2 $$
we find
$$ S_8 = x^2 y^2 (x^2 - z^2)^2 + \text{ cyclic } + \frac12 x^4 (y^2 - z^2)^2 + \text{ cyclic } $$