Alan

Correct. The solutions are irrational, the coefficients of the cubic polynomial are rational. 

I guess I shouldn't have put the emphasis on the word "irrational" in my earlier reply.  However, I think the terms with the surds should come in  complementary pairs.  For example, the cubic can be rewritten as follows:

All solutions are (I suspect) irrational, but two are in the form a + b√3 and a - b√3.

True!  Perhaps I just got lucky!

Exponentiation takes priority over negation, so -2^4 is -(2^4).  If you want the negation to take priority you would need to emphasise this by writing (-2)^4.

"If three real numbers are in an arithmetic progression while their squares are in a geometric progression, find the sum of all the possible ratios of the geometric progression of the squares."

The question specified that the polynomial had rational coefficients.  The only way that could occur here is if the irrational terms in sqrt(3) and sqrt(2) were eliminated by complementary terms.  In general, an n'th order polynomial will have n roots (counting possible repeated ones).  Therefore with an odd degree polynomial with rational coefficients you couldn't have an odd number of roots that all contained irrational terms; the irrational ones would have to appear in complementary pairs. 

Like so:

My attempt is as follows:

Edit: common ratio should be lambda^-1; the RHS of the second equation on the line beginning "hence" should have a 1 in the numerator not a lambda.

As a check, here is WolframAlpha's factorisation:

See https://web2.0calc.com/questions/help_40754#r2

We have yet another annoying situation where someone posts the question twice instead of continuing the original post!