If \(x, y\) are reals with \(x+y=10\) and \(x^3-y^3=504\), what is \(x-y\)?
+1312 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.
Hmm... I see what you did there, but I still (embarrassingly) have no clue of what to do next.
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.
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.
