Ratio problem 2 by Dhaval Furia

Let the two positive numbers be $a$ and $b$ , such that $a>b$ . Then $ab =616$ and

$\begin{aligned} \frac{a^3-b^3}{(a-b)^3} & = \frac {157}3 \\ 3(a^3-b^3) & = 157(a-b)^3 \\ 3(a-b)(a^2+ab+b^2) & = 157(a-b)^3 \\ 3(a^2+ab+b^2) & = 157(a^2-2ab+b^2) \\ 3(a^2+2ab+b^2) - 3ab & = 157(a^2+2ab+b^2) - 628ab \\ 3(a+b)^2 - 3ab & = 157(a+b)^2 - 628ab \\ 154(a+b)^2 & = 625ab \\ (a+b)^2 & = \frac {625 \cdot 616}{154} = 2500 \\ \implies a+b & = \boxed{50} \end{aligned}$

Let the numbers be $a$ and $b$ . Then $ab=616$

$\dfrac{a^3-b^3}{(a-b) ^3}=\dfrac{157}{3}$

Solving, we get $a^2+b^2=1268\implies (a+b)^2=a^2+b^2+2ab=1268+1232=2500\implies a+b=\boxed {50}$ .