Motzkin's original proof shows that x^4y^2 + x^2y^4 + z^6 - a x^2y^2z^2 is psd and
not sos for any a in the interval (0,3]. If you take a = .02 say, it is reasonably
simple, though messy, to show that (x^4y^2 + x^2y^4 + z^6 - .02x^2y^2z^2)^3 is a 
sum of squares; in fact, it's a sum of binomial squares (x^b y^c z^d - x^e y^f z^g)^2, 
where b+c+d=e+f+g=9. The idea is to look at any monomial with a negative coefficient
and make it into the middle term of this square, in a way that the other two terms
are still in the Newton polytope. For example, one term in the given cube is -.06x^10y^6z^2, which is "handled" by .03(x^6y^3 - x^4y^3z^2)^2. It's sort of messy to work out, but I've convinced myself (at least) that it's true.