Results 1 to 4 of 4

Math Help - [SOLVED] Difference of cubes

  1. #1
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    Question [SOLVED] Difference of cubes

    x^3+y^3=208
    x and y are natural numbers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by courteous View Post
    x^3+y^3=208
    x and y are natural numbers.
    That's the sum of cubes

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

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

    Since x and y are natural numbers, x+y>0 and 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...

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

    Note that 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 >.>
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member courteous's Avatar
    Joined
    Aug 2008
    From
    big slice of heaven
    Posts
    206

    Red face Corrected - it is a minus...

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

    The correct(ed) version:
    Quote Originally Posted by courteous View Post
    x^3-y^3=208
    x and y are natural numbers.
    The solution is 6^3-2^3=208. But how do you get it?
    -----------------------------------------------
    Moo, about developing from (x+y)^3: what do you do with 3x^2y and 3xy^2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Moo, about developing from (x+y)^3: what do you do with 3x^2y and 3xy^2?
    Factor 3xy and see what's left

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

    The correct(ed) version:


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

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

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

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

    Exactly the same reasoning for the other possibilities !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The Sum and difference of cubes
    Posted in the Algebra Forum
    Replies: 11
    Last Post: January 20th 2010, 08:16 PM
  2. Sum or difference of cubes.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 14th 2010, 06:52 PM
  3. How to solve Difference of Cubes?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 13th 2009, 09:48 PM
  4. Difference of cubes
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 2nd 2009, 08:39 PM
  5. Difference of Cubes
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 21st 2007, 05:21 AM

Search Tags


/mathhelpforum @mathhelpforum