Question

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

• A. 58
• B. 50
• C. 95
• D. 85

Answer 1

The Correct Option is B

Given that the product of two numbers is: \[ x \cdot y = 616 \] and, \[ \frac{x^3 - y^3}{(x - y)^3} = \frac{157}{3} \]

Assume:

• \( x^3 - y^3 = 157k \)
• \( (x - y)^3 = 3k \)

We use the identity:

\[ (x - y)^3 = x^3 - y^3 - 3xy(x - y) \]

Substitute the expressions:

\[ 3k = 157k - 3 \cdot 616 \cdot (x - y) \]

Rewriting and solving:

\[ 154k = 3 \cdot 616 \cdot (x - y) \]

Let \( x - y = (3k)^{1/3} \). Then:

\[ 154k = 3 \cdot 616 \cdot (3k)^{1/3} \]

Divide both sides by 154:

\[ k = 12 \cdot (3k)^{1/3} \]

Cube both sides:

\[ k^3 = 12^3 \cdot 3k \Rightarrow k^2 = 3 \cdot 12^3 \Rightarrow k = 72 \]

Now, compute \( x - y \):

\[ x - y = (3k)^{1/3} = (3 \cdot 72)^{1/3} = 216^{1/3} = 6 \]

Use the identity:

\[ (x + y)^2 = (x - y)^2 + 4xy = 6^2 + 4 \cdot 616 = 36 + 2464 = 2500 \]

\[ x + y = \sqrt{2500} = 50 \]

The sum of the two numbers is: \(\boxed{50}\)