[tex]\text{We have:}\\\\f(x)=\dfrac{a}{x^3}\to y=\dfrac{a}{x^3}\\\\\text{and the point}\ \left(2,\ \dfrac{5}{4}\right)\to x=2,\ y=\dfrac{5}{4}.\\\\\text{Substitute:}\\\\\dfrac{5}{4}=\dfrac{a}{2^3}\\\\\dfrac{5}{4}=\dfrac{a}{8}\qquad\text{multiply both sides by 8}\\\\a=\dfrac{5}{\not4_1}\cdot\not8^2\\\\\boxed{a=10}[/tex]
