Since, $\alpha ,\beta$ be the roots of the equation $x^2+{60}^{\frac{1}{4}}x+a=0$, therefore

$\alpha +\beta =-{60}^{\frac{1}{4}}$

$\alpha \beta =a$

Given:

${\alpha }^4+{\beta }^4=-30$

$\Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^2-2{\alpha }^2{\beta }^2=-30$

$\Rightarrow {\left\{{\left(\alpha +\beta \right)}^2-2\alpha \beta \right\}}^2-2a^2=-30$

$\Rightarrow {\left\{{60}^{\frac{1}{2}}-2a\right\}}^2-2a^2=-30$

$\Rightarrow 60+4a^2-4a\times {60}^{\frac{1}{2}}-2a^2=-30$

$\Rightarrow 2a^2-4·{60}^{\frac{1}{2}}a+90=0$

$\Rightarrow a^2-2·{60}^{\frac{1}{2}}a+45=0$

Product of roots$=45$