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Question 53
The sum of the cubes of two numbers is 128, while the sum of the reciprocals of their cubes is 2.
What is the product of the squares of the numbers?
Solution
Considering the two numbers to a, b :
We were given that :
$$a^3+b^3\ =\ 128$$
$$\frac{1}{a^3}+\ \frac{1}{b^3}=\ 2$$
$$\frac{\left(a^3+b^3\right)}{a^3\cdot b^3}=\ 2\ =\ \frac{128}{k}$$
k = 64.
Hence $$a^3\cdot b^3\ =\ 64$$
a*b = 4 and $$a^2\cdot b^2\ =\ 16$$
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