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anonymous

  • 5 years ago

x^6-7x^3-8=0

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  1. radar
    • 5 years ago
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    I am assuming that you want to simplify this. It will simplify first factor it like this: \[(x ^{3}+1)(x ^{3}-8)\] The second factor is the difference of two perfect cubes and can be factored further.

  2. radar
    • 5 years ago
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    this can be factored further as one is the sum of two perfect cubes, and the other is the difference. I will do them one at a time and then combine

  3. anonymous
    • 5 years ago
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    ok

  4. anonymous
    • 5 years ago
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    i have that just dont know where to go from there

  5. radar
    • 5 years ago
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    \[(x ^{3}=1)=(x+1)(x ^{2}-x+1)\]

  6. radar
    • 5 years ago
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    That should be (x^3+1) = lol The other factor is the difference of two perfect cubes \[(x ^{3}-8)=(x-2)(x ^{2}+2x+4)\] Now the total factors can be expressed as:\[(x-2)(x+1)(x ^{2}-x+1)(x ^{2}+2x+4)\]

  7. anonymous
    • 5 years ago
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    7/2 (7/2)^2=49/4 x^6-(7/2)^3=8+49/4 (x^3-(7/2))^2=81/4 x^3-(7/2)=+ or - sqrt{81/4} x^3=(7/2) + or - sqrt{81/4} x^3=(7/2) + or - sqrt{(81*4)/4} x^3=(7/2) + or - sqrt{324} x^3=(7/2) + or - 9/2 x^3=(7/2)+(9/2) x^3=8 x=2^3 x=2 or: x^3=(7/2)-(9/2) x^3=-2/2 x=sqrt[3]{-1} x=-1

  8. radar
    • 5 years ago
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    Don't know how that got in there, seems to be an answer to a different problem

  9. radar
    • 5 years ago
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    The web is acting crazy, can't seem to enter anything! Hope you got it all

  10. radar
    • 5 years ago
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    I refreshed my browser and it appears ok now. Did you understand the final factors?

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