Let $$a \in R$$ and let $$\alpha, \beta$$ be the roots of the equation $$x^2 + 60^{1/4}x + a = 0$$. If $$\alpha^4 + \beta^4 = -30$$, then the product of all possible values of $$a$$ is _____.

We need to find the product of all possible values of $$a$$ given that $$\alpha, \beta$$ are roots of $$x^2 + 60^{1/4}x + a = 0$$ and $$\alpha^4 + \beta^4 = -30$$.

By Vieta’s formulas for the equation $$x^2 + 60^{1/4}x + a = 0$$, we have $$\alpha + \beta = -60^{1/4}$$ and $$\alpha\beta = a$$.

To express $$\alpha^4 + \beta^4$$ in terms of $$a$$, note that
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \sqrt{60} - 2a$$
and
$$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (\sqrt{60} - 2a)^2 - 2a^2 = 60 - 4a\sqrt{60} + 4a^2 - 2a^2 = 2a^2 - 4a\sqrt{60} + 60$$.

Since $$\alpha^4 + \beta^4 = -30$$, we set $$2a^2 - 4a\sqrt{60} + 60 = -30$$, which simplifies to $$2a^2 - 4a\sqrt{60} + 90 = 0$$ and then to $$a^2 - 2\sqrt{60}\,a + 45 = 0$$.

For the quadratic $$a^2 - 2\sqrt{60}\,a + 45 = 0$$, Vieta’s formulas show that the product of the roots is $$\dfrac{45}{1} = 45$$.

The required product of all possible values of $$a$$ is $$\boxed{45}$$.