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anonymous

  • one year ago

how to factor this ? 1+y^12 =

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  1. anonymous
    • one year ago
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    that is a sum of cubes \[a^3+b^3=(a+b)(a^2-ab+b^2)\] Where in your expression \(a=1\) and \(b=y^4\)

  2. anonymous
    • one year ago
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    thanks :)

  3. anonymous
    • one year ago
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    @peachpi how to determine faster if square or cube ?

  4. anonymous
    • one year ago
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    For a difference I'm not sure it matters because you will end up with the same solution either way. For a sum, you can only factor a sum of cubes with involving complex numbers.

  5. anonymous
    • one year ago
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    the exponent of the variable ? like 12 ? how do we know faster to determine square or cube ? how about like y^18 y^24 ?

  6. anonymous
    • one year ago
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    For exponents divisible by both 2 and 3 the exponents themselves don't matter, it's the sign between the terms that does. 1+y^12, 1+y^18 and 1+y^24 you have to factor using cube because there's no rule for a sum of squares. 1-y^12, 1 - y^18, and 1 - y^24 I believe you should be able to apply either the difference of cubes or difference of squares rule and get the same result. I personally prefer to do the square factor first because then you don't have to deal with the trinomial square first 1 - y^12 (1 + y^6)(1-y^6) For (1 + y^6) you have to use the sum of cube rule. For (1-y^6) you can again choose either rule. I'll do square to be consistent (1 + y^2)(1 - y^2 + y^4)(1+y^3)(1-y^3) The first two are irreducible, the second two are no longer squares, so they must be factored using as cubes (1 + y^2)(1 - y^2 + y^4)(1+y)(1 - y + y^2)(1 - y)(1 + y + y^2) Try doing the cube first and you should get the same result

  7. anonymous
    • one year ago
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    thanks for your time :)

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