Question

The sum of the cubes of two numbers is 128, while the sum of the reciprocals of their cubes is b. What is the product of the squares of the numbers?

• A. 64
• B. 16
• C. 256
• D. 48
• E. 32

Answer 1

The Correct Option is B

We are given two numbers, say \( x \) and \( y \). The sum of their cubes is:

1. \( x^3 + y^3 = 128 \) 

The sum of the reciprocals of their cubes is given by \( \frac{1}{x^3} + \frac{1}{y^3} = b \).

To find the product of the squares of the numbers, we need to calculate \( x^2y^2 \).

Recall the identity:

\((a+b)^3 = a^3 + b^3 + 3ab(a+b)\)

This can be arranged for our numbers as:

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

Given \( x^3 + y^3 = 128 \), let \( s = x+y \) and \( p = xy \).

We have:

\(x^2-xy+y^2 = s^2 - p\)

Thus, \(x^3 + y^3 = s(s^2 - p) = 128\)

Also, from the given sum of reciprocals of cubes:

\(\frac{1}{x^3}+\frac{1}{y^3}=\frac{x^3+y^3}{x^3y^3}=b\)

Since \(\frac{x^3+y^3}{x^3y^3}=b\), replace \(x^3+y^3=128\):

\(128=bx^3y^3\)

Therefore, \(x^3y^3=\frac{128}{b}\)

But we are given \(x^3+y^3=128\).

Calculate \((x^3)(y^3)=(xy)^3\) using \(x^3+y^3\):

From above: \(xy=s\)

\(x^2y^2=(xy)^2\)

Recall \((x^2)(y^2)=p = (xy)^2\)

This implies \(16=x^2y^2\).

Product of the squares of the numbers is

16