The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

The correct option is 2 (which corresponds to 50).

Let the two positive numbers be $x$ and $y$, with $x>y>0$.

We are given that their product is:
$xy=616$

We are also given the ratio of the difference of their cubes to the cube of their difference is 157:3:
$\frac{x^3-y^3}{{(x-y)}^3}=\frac{157}{3}$

Recall the algebraic identities:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
and
${(x-y)}^3=(x-y){(x-y)}^2$

Substituting these into the ratio equation, we can cancel $(x-y)$ since the numbers are positive and distinct (as their ratio involves a non-zero difference):
$\frac{x^2+xy+y^2}{{(x-y)}^2}=\frac{157}{3}$

We can rewrite the numerator $x^2+xy+y^2$ as ${(x-y)}^2+3xy$:
$\frac{{(x-y)}^2+3xy}{{(x-y)}^2}=\frac{157}{3}$

Dividing the terms on the left side, we get:
$1+\frac{3xy}{{(x-y)}^2}=\frac{157}{3}$

Subtract 1 from both sides:
$\frac{3xy}{{(x-y)}^2}=\frac{154}{3}$

Substitute $xy=616$ into the equation:
$\frac{3\times 616}{{(x-y)}^2}=\frac{154}{3}$

Simplify the equation by dividing both sides by 154 (note that $616=4\times 154$):
$\frac{3\times 4}{{(x-y)}^2}=\frac{1}{3}$

Cross-multiplying gives:
${(x-y)}^2=3\times 4\times 3=36$

Since $x>y$, we have:
$x-y=6$

We need to find the sum of the two numbers, $x+y$. We can use the relation:
${(x+y)}^2={(x-y)}^2+4xy$

Substitute the values of ${(x-y)}^2$ and $xy$:
${(x+y)}^2=36+4\times 616$
${(x+y)}^2=36+2464=2500$

Taking the positive square root (since $x$ and $y$ are positive):
$x+y=50$

Thus, the sum of the two numbers is 50.