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August899

  • one year ago

Factor the polynomials:m^3+1/8 and p^3+216

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  1. zepdrix
    • one year ago
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    Hey :)

  2. zepdrix
    • one year ago
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    \[\large\rm m^3+\frac{1}{8}\]Hmmm, recall that we have a rule for factoring the `sum of cubes`: \[\large\rm a^3+b^2=(a+b)(a^2-ab+b^2)\]

  3. zepdrix
    • one year ago
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    But with the way our problem is written right now, we only have one cube. Any ideas how to turn the 1/8 into `something`^3 ?

  4. zepdrix
    • one year ago
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    @August899

  5. August899
    • one year ago
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    do we divide 1 over 8?

  6. zepdrix
    • one year ago
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    The 8 is some number to the third power.

  7. zepdrix
    • one year ago
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    Example: 27 is 3 to the third power. Because 3 x 3 x 3 = 27 Therefore \(\large\rm 27=3^3\)

  8. zepdrix
    • one year ago
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    Comeon Auguuuuuust >.< break down the 8, how can we write that another way?

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