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Math Help - Absurdly Complicated (to me) Factoring Problem

  1. #1
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    Absurdly Complicated (to me) Factoring Problem

    I was given a review handout for a Calc. + Anal. Geom. (lol!) and I ran into an inconspicuously confusing factoring problem:

    54ał - 16bł

    I factored it to 2(27ał - 8bł), but he got to
    2(3a - 2b)(9a˛ + 6ab + 4b˛)

    May someone please explain to me how solving this this was expected to be innate algebraic practicality?

    …or please explain how he might have gotten to his solution. From there, I hope I'd be able to tackle almost any similarly ridonque. factoring problem…
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  2. #2
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    Quote Originally Posted by leveebreaker05 View Post
    I was given a review handout for a Calc. + Anal. Geom. (lol!) and I ran into an inconspicuously confusing factoring problem:

    54ał - 16bł

    I factored it to 2(27ał - 8bł), but he got to
    2(3a - 2b)(9a˛ + 6ab + 4b˛)

    May someone please explain to me how solving this this was expected to be innate algebraic practicality?

    …or please explain how he might have gotten to his solution. From there, I hope I'd be able to tackle almost any similarly ridonque. factoring problem…
    pattern for factoring the difference of cubes ...

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

    in your problem, x = 3a and y = 2b
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  3. #3
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    More generally, it is true that
    x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot\+ x^2y^{n- 3}+ xy^{n-2}+ y^{n-1})
    and, if n is odd, that
    x^n+ y^n= (x+ y)(x^{n-1}- x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot\- x^2y^{n- 3}+ xy^{n-2}- y^{n-1})
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  4. #4
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    Talking

    Quote Originally Posted by leveebreaker05 View Post
    May someone please explain to me how...he might have gotten to his solution.
    He applied the formula for factoring a difference of cubes. This is a formula that you should memorize!
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  5. #5
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    In my opinion,

    54(a^3)-16(b^3) = 2[27(a^3)-8(b^3)]

    Then, we notice that 27^(1/3)=3, and -8^(1/3)=-2.
    Then,
    (3a-2b)^3 = 27(a^3)-8(b^3)-54(a^2)(b)+36(a)(b^2)
    27(a^3)-8(b^3)=(3a-2b)^3+54(a^2)(b)-36(a)(b^2)

    Then,
    54(a^3)-16(b^3) = 2[(3a-2b)^3+54(a^2)(b)-36(a)(b^2)]
    = 2[(3a-2b)^3+18ab(3a-2b)]
    = 2[(3a-2b)((3a-2b)^2+18ab)]
    = 2(3a-2b)(9(a^2)+6ab+4(b^2))
    That's the answer....but i think maybe got some better solution...
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