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

Let us call our two positive numbers $x$ and $y$.

Given Information:

Product: $xy = 616$

Ratio: $\dfrac{x^3 - y^3}{(x - y)^3} = \dfrac{157}{3}$

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We need to simplify $x^3 - y^3$. There's a useful algebraic identity here:

Key Identity: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Why does this work? If you expand $(x - y)(x^2 + xy + y^2)$:

$(x - y)(x^2 + xy + y^2) = x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3$

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Now our ratio becomes:

$\dfrac{x^3 - y^3}{(x - y)^3} = \dfrac{(x - y)(x^2 + xy + y^2)}{(x - y)^3}$

The $(x - y)$ terms cancel out:

$\dfrac{(x - y)(x^2 + xy + y^2)}{(x - y)^3} = \dfrac{x^2 + xy + y^2}{(x - y)^2}$

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We need to expand $(x - y)^2$:

$(x - y)^2 = x^2 - 2xy + y^2$

So our equation becomes:

$\dfrac{x^2 + xy + y^2}{x^2 - 2xy + y^2} = \dfrac{157}{3}$

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Since $xy = 616$, we can substitute:

$\dfrac{x^2 + y^2 + 616}{x^2 + y^2 - 2(616)} = \dfrac{157}{3}$

$\dfrac{x^2 + y^2 + 616}{x^2 + y^2 - 1232} = \dfrac{157}{3}$

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Let $t = x^2 + y^2$ to make calculations easier:

$\dfrac{t + 616}{t - 1232} = \dfrac{157}{3}$

Cross-multiplying:

$3(t + 616) = 157(t - 1232)$

$3t + 1848 = 157t - 193424$

$1848 + 193424 = 157t - 3t$

$195272 = 154t$

$t = \dfrac{195272}{154} = 1268$

Therefore: $x^2 + y^2 = 1268$

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Key Identity: $(x + y)^2 = x^2 + 2xy + y^2$

We know:

$x^2 + y^2 = 1268$

$xy = 616$

Substituting:

$(x + y)^2 = 1268 + 2(616) = 1268 + 1232 = 2500$

Taking the square root:

$x + y = \sqrt{2500} = 50$

We take the positive root since both numbers are positive.

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Answer: The sum of the two numbers is 50.