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anonymous

  • 5 years ago

Ok one more 16s^6-2g^6 so 2s(2s)^2-2g(g)^2

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  1. anonymous
    • 5 years ago
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    hmmm....is that your simplification?

  2. anonymous
    • 5 years ago
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    Yes am I wrong is this not the answer?

  3. anonymous
    • 5 years ago
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    yeah you have to notice that (2s)^2 = 4s^2 and so multiplying that by 2s would be 8s^3 and 2g*g^2 = 2g^3

  4. anonymous
    • 5 years ago
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    Then am I to assume that (2s)^4 (2g)3

  5. anonymous
    • 5 years ago
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    no because then you have 16s^4

  6. anonymous
    • 5 years ago
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    one way to look at it is seperating the equation into a product of binomials

  7. anonymous
    • 5 years ago
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    ok

  8. anonymous
    • 5 years ago
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    hold on let me write something up first on paper to see if its correct

  9. anonymous
    • 5 years ago
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    ok i got something

  10. anonymous
    • 5 years ago
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    2s(2s)^5-2g(g)^5

  11. anonymous
    • 5 years ago
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    so notice factoring out a 2 you get \[2( 8s ^{6} - g^{6})\] and doing a little algebra we get \[2( (2s^{2})^{3} - (g^{2})^{3})\] and so using the difference of cubes we get \[2[ (2s^{2} - g^{2})* ( (2s^{2})^{2} +2s^{2}g^{2} + (g^{2})^{2})\]

  12. anonymous
    • 5 years ago
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    what you did is incorrect since (2s)^5 = 32s^5 and multiplying by 2s would be 64s^6

  13. anonymous
    • 5 years ago
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    How about: \[-2(g^2-2s^2)(g^4+2g^2 s^2+4s^4)\]

  14. anonymous
    • 5 years ago
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    dude if you look at what i wrote what you wrote is the exact same

  15. anonymous
    • 5 years ago
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    Yes, after removing the open bracket symbol they are equivalent. Good work robotsandtoast.

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