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Math Help - factorizing

  1. #1
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    factorizing

    factorize (x+3y)^3-x^3-27y^3
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    Quote Originally Posted by requal View Post
    factorize (x+3y)^3-x^3-27y^3
    I suggest that you first expand the perfect cube and then simplify the overall expression. You will be left with two terms, which you can factorise by taking out a common factor.
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    (x+3y)^3-x^3-27y^3=9 x y (x+3 y)
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    Hello,

    Use this identity :
    a^3+b^3=(a+b)(a^2-ab+b^2)

    We here have :
    (x+3y)^3-(x^3+27y^3)=(x+3y)^3-(x^3+(3y)^3)=(x+3y)^3-(x+3y)(x^2-3xy+9y^2)

    =(x+3y) \left((x+3y)^2-(x^2-3xy+9y^2)\right)

    =(x+3y) (x^2+6xy+9y^2-x^2+3xy-9y^2)

    =(x+3y)(9xy)
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