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
- 58
- 50
- 95
- 85
The Correct Option is B
Solution and Explanation
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: \(\boxed{50}\)
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