In the equation $x^2 - \frac{15}{4}x + a^3 = 0$, one of the roots is the square of the other if $a$ is equal to

Let the roots be $r$ and $r^2$. Then by Vieta's formulas, their sum is $$r+r^2=\frac{15}{4}$$ and their product is $$r^3=a^3.$$ From the sum, $$r^2+r-\frac{15}{4}=0.$$ Multiplying by $4$, $$4r^2+4r-15=0.$$ Factoring, $$(2r-3)(2r+5)=0.$$ So $$r=\frac{3}{2} \text{ or } r=-\frac{5}{2}.$$ Hence $$a^3=r^3,$$ so $$a=r=\frac{3}{2} \text{ or } a=-\frac{5}{2}.$$ This matches option $\mathrm{D}$.