SOLUTION: The sum of cube root of two numbers is 128,while the sum of the reciprocals of their cubes is 2. What are thosw two numbers...Thank You..

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Question 1198694: The sum of cube root of two numbers is 128,while the sum of the reciprocals of their cubes is 2. What are thosw two numbers...Thank You..
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Given,
Cube root of x + Cube root of y = 128
x%5E%281%2F3%29%2By%5E%281%2F3%29=128 .................(1)
Also,
1%2F%28x%5E3%29+%2B+1%2F%28y%5E3%29+=+2 .......(2)
From equation (1), we can write,
y^(1/3) = 128 - x^(1/3)
Now replace y^(1/3) in second equation,
1%2Fx%5E3+%2B+1%2F%28%28128+-+x%5E%281%2F3%29%29%29%5E3=+2
On simplification we get cubic equation in x^(1/3)
3 * x^(1/3)^4 - 8192 * (x^(1/3)) + 2097152 = 0
On solving this equation, we get x^(1/3)= 32
Now replace this value in equation (1),
y^(1/3) = 128 - 32 = 96
numbers are x = 32^3 = 32768 and y = 96^3 = 884736.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

The numbers that @mananth announced as solutions

        x = 32^3,   y = 96^3

do not satisfy equation

        1%2Fx%5E3 + 1%2Fy%5E3 = 2.

Therefore,  what he calls  " the solution ",  is not a solution, at all.

It looks like he is unfamiliar with the conception of checking a solution, at all.


Absolutely mathematically illiterate person.

But wants very much to teach others.