Results 1 to 12 of 12

Math Help - The Sum and difference of cubes

  1. #1
    Member
    Joined
    Jan 2009
    From
    Ontario Canada
    Posts
    243

    The Sum and difference of cubes

    Hello,
    Need help in these problems

    1-(x+y)^3

    (x+y)^3+1

    1-a^6x^6

    I am having some trouble figuring these out, if you can help me please, thanks.

    If you can key it out fully would be great to help me understand.
    Thank you
    Joanne
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by bradycat View Post
    Hello,
    Need help in these problems

    1-(x+y)^3
    you have 1^3 - (x + y)^3

    apply the difference of two cubes formula: a^3 - b^3 = (a - b)(a^2 + ab  + b^2)

    here a = 1, b = x + y

    try it

    (x+y)^3+1
    formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2)

    here, a = x + y, b = 1

    1-a^6x^6
    write it as 1 - [(ax)^3]^2.

    applying the difference of two squares formula, we get: [1 + (ax)^3][1 - (ax)^3]

    now, apply the sum of two cubes formula to the first factor, and the difference of two cubes formula to the last factor
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,139
    Thanks
    1013
    Quote Originally Posted by bradycat View Post
    Hello,
    Need help in these problems

    1-(x+y)^3

    (x+y)^3+1

    1-a^6x^6

    I am having some trouble figuring these out, if you can help me please, thanks.

    If you can key it out fully would be great to help me understand.
    Thank you
    Joanne
    note the factoring patterns for the sum and difference of two cubes ...

    a^3 + b^3 = (a+b)(a^2 - ab + b^2)

    a^3 - b^3 = (a-b)(a^2 + ab + b^2)

    for the first problem ...

    a is 1

    b is (x+y)

    1^3 - (x+y)^3 = [1 - (x+y)][1^2 + 1(x+y) + (x+y)^2]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    use binomial theorem
    it is ussually in the book cover or listed in the reference

    (x+y)^3 = x^3+3x^2y+3xy^2+y^3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    From
    Ontario Canada
    Posts
    243

    Hello Again and thanks for the reply.

    Answers to first one is
    1-(x+y)^3 = (1-x-y)(1+x+y+(x+y)^2 which I got now, thanks.

    (x+y)^3+1 = I can't find the full answer to this, sorry.

    1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
    This one I am still stuck on doing. I must be doing something wrong, and can't figure it out.
    Can someone completely display this answer, thanks.
    Joanne
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,139
    Thanks
    1013
    Quote Originally Posted by bradycat View Post
    (x+y)^3+1 = I can't find the full answer to this, sorry.
    [(x+y)+1][(x+y)^2 - (x+y)(1) + 1^2]

    clean it up.


    for 1-a^6x^6 , try factoring ...

    1^3 - (a^2x^2)^3
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    theorum

    (x+y)^3 = x^3+3x^2y+3xy^2+y^3

    so just add 1 to it

    (x+y)^3+1 = x^3+3x^2y+3xy^2+y^3+1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by bradycat View Post
    (x+y)^3+1 = I can't find the full answer to this, sorry.
    skeeter dealt with this

    1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
    This one I am still stuck on doing. I must be doing something wrong, and can't figure it out.
    Can someone completely display this answer, thanks.
    Joanne
    what you have here is correct
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by bigwave View Post
    use binomial theorem
    it is ussually in the book cover or listed in the reference

    (x+y)^3 = x^3+3x^2y+3xy^2+y^3
    Quote Originally Posted by bigwave View Post
    theorum

    (x+y)^3 = x^3+3x^2y+3xy^2+y^3

    so just add 1 to it

    (x+y)^3+1 = x^3+3x^2y+3xy^2+y^3+1
    well, the OP was a little vague, but judging from the title, i think the idea here is to factor, not expand.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jan 2009
    From
    Ontario Canada
    Posts
    243
    1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)

    That is the answer and I am trying this and still not getting anywhere.
    I get (1-a^2x^2) (1^2+ax+a^4x^4)
    So then you get(1-ax)(1+ax) but stuck on this if it's suppose to look like the answer above.
    HELP PLEASE.
    thks
    Jo
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jan 2009
    From
    Ontario Canada
    Posts
    243

    Skeeter

    The reply you have above saying I have it right. That is the answer in my work book. I am stuck on getting the last part of the problem as I noted above.
    Jo
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by bradycat View Post
    The reply you have above saying I have it right. That is the answer in my work book. I am stuck on getting the last part of the problem as I noted above.
    Jo
    i told you how to get to that answer in my first post. apply the difference of two squares first, then the sum/difference of two cubes.

    if you apply the difference of two cubes first (which makes the problem harder) you would get [1 - (ax)^2][1 + (ax)^2 + (ax)^4]

    you would then need to factor 1 - (ax)^2 using the difference of two squares, and to factor 1 + (ax)^2 + (ax)^4, you would need to write it as (ax)^4 + 2(ax)^2 + 1 - (ax)^2 = [(ax)^2 + 1]^2 - (ax)^2 and use the difference of two sqaures again. See? too much work. review my first post and follow the suggestion there
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum or difference of cubes.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 14th 2010, 06:52 PM
  2. Simplifying: difference of two cubes
    Posted in the Algebra Forum
    Replies: 10
    Last Post: April 8th 2009, 06:10 AM
  3. Difference of cubes
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 2nd 2009, 08:39 PM
  4. [SOLVED] Difference of cubes
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 13th 2008, 07:26 AM
  5. Difference of Cubes
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 21st 2007, 05:21 AM

Search Tags


/mathhelpforum @mathhelpforum