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If \(x, y\) are reals with \(x+y=10\) and \(x^3-y^3=504\), what is \(x-y\)?

 Oct 3, 2022
 #1
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Difference of cubes: (x - y)(x^2 + xy + y^2)

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

(x - y)((x + y)^2 - xy)

\((x - y)(100 - xy) = 504\)

 

How can you take it from here? Good luck.

 Oct 3, 2022
 #2
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Hmm... I see what you did there, but I still (embarrassingly) have no clue of what to do next. 

 Oct 3, 2022
 #3
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No, I also don't see it.

It's a dissapointing question in a way, in that it's easy to spot the (or a) solution.

We are not told whether or not x and y are integers or whether or not they are both positive, but assuming they are both positive integers, the smallest possible value for x is 8.

Now, 8 cubed = 512 and 512 - 8 = 504.

Since 8 = 2 cubed, that gets you the solution x = 8, y = 2.

 Oct 4, 2022
 #4
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Oh! Thank you!

 Oct 4, 2022
 #5
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Here's an algebraic routine that gets you to the complete solution.

Let x = u + v and y = u - v, so that u = (x + y)/2 and v = (x - y)/2.

Then,

\(\displaystyle x^{3} - y^{3} = (u + v)^{3} - (u - v)^{3} \\ = u^{3}+3u^{2}v+3uv^{2}+v^{3}-(u^{3}-3u^{2}v+3uv^{2}-v^{3})\\ =6u^{2}v +2v^{2}.\)

\(\displaystyle u^{2} = (x+y)^{2}/4=100/4=25,\)

so

\(\displaystyle 6.25.v+2v^{3}=504,\\ v^{3}+75v-252 = 0.\)

That has \(v=3\)

as an obvious solution from which

\(\displaystyle (v-3)(v^{2} +3v+84)=0.\)

So,

 \(v=(x-y)/2=3,\\ x-y=6.\)

The other two possibles are complex.

 Oct 4, 2022

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