# Thread: [SOLVED] Difference of cubes

1. ## [SOLVED] Difference of cubes

$\displaystyle x^3+y^3=208$
x and y are natural numbers.

2. Hello,
Originally Posted by courteous
$\displaystyle x^3+y^3=208$
x and y are natural numbers.
That's the sum of cubes

Well, well... The formula is $\displaystyle x^3+y^3=(x+y)(x^2-xy+y^2)$
If you cant remember it, start from $\displaystyle (x+y)^3$ and develop it

Solving for it... ? Not impossible.
Note that $\displaystyle 208=2^4*13$

Since x and y are natural numbers, $\displaystyle x+y>0$ and $\displaystyle x^2-xy+y^2>0$ are integers.

So the two factors will be a combination of the divisors of 208. For example 26 and 8 or 13 and 16 or...

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Try all the possible values for x+y among 1,2,4,8,13,16,26,52,104,208 ()
Then note that $\displaystyle x^2-xy+y^2=(x+y)^2-3xy$ and is respectively among 208,104,...,2,1

Note that $\displaystyle x+y \neq 1,2,4$ otherwise the product can't be large enough (check it )

If you need more explanations, do tell me. But as far as I can see, it will be a long way to do >.>

3. ## Corrected - it is a minus...

I've correctly titled the thread, but have mistakenly put + (plus) instead of - (minus).

The correct(ed) version:
Originally Posted by courteous
$\displaystyle x^3-y^3=208$
x and y are natural numbers.
The solution is $\displaystyle 6^3-2^3=208$. But how do you get it?
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Moo, about developing from $\displaystyle (x+y)^3$: what do you do with $\displaystyle 3x^2y$ and $\displaystyle 3xy^2$?

4. Moo, about developing from $\displaystyle (x+y)^3$: what do you do with $\displaystyle 3x^2y$ and $\displaystyle 3xy^2$?
Factor $\displaystyle 3xy$ and see what's left

Originally Posted by courteous
I've correctly titled the thread, but have mistakenly put + (plus) instead of - (minus).

The correct(ed) version:

The solution is $\displaystyle 6^3-2^3=208$. But how do you get it?
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Develop (x-y)^3 to get $\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)$

So we know that x and y are positive -----> $\displaystyle x^2+xy+y^2 > 0$ and $\displaystyle x-y>0$

$\displaystyle x^2+xy+y^2=(x-y)^2+3xy$
Obviously, $\displaystyle x^2+xy+y^2>x-y$

$\displaystyle \begin{array}{c|cc} x-y & x^2+xy+y^2 \\ \hline 1&208& \\ 2&104& \\ 4&52& \\ 8&26& \\ 13&16& \\ 16&13& \rightarrow \text{hey ! } x-y> x^2+xy+y^2 \text{ ? Impossible !} \end{array}$

So you have 5 possibilities to test out (x-y stopping at 13).

For example. If $\displaystyle x-y=1$ then $\displaystyle x^2+xy+y^2=(x-y)^2+3xy=1^2+3xy$
Are there x and y such that $\displaystyle 1+3xy=208$ and $\displaystyle x-y=1$ ? That's a linear system of equations

Exactly the same reasoning for the other possibilities !