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anonymous

  • one year ago

HELP!! Can anyone solve this?? 1) x^8 - y^8 2) 9x^4 - 16y^8 3) 8x^6 - 27 4) 4x^2/9 - y^4 * Difference of Squares & Sum or Difference of Cubes

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  1. anonymous
    • one year ago
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    1.) x^8 - y^8 = (x4 + y4) • (x4 - y4) - squares 2.) \[3^2x^4 - 2^4y^8\] - squares 3.) 8x^6-27 - cubes 4.) I'm not sure on this one. Is 2/9 a fraction or is it dividing the term 4x^2 by 9? If I had to guess, it would be squares.

  2. anonymous
    • one year ago
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    Thank you for your answering and... yes it is divided by 9 in no. 4

  3. anonymous
    • one year ago
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    is it really minus in squares on no. 1??? Im confused

  4. freckles
    • one year ago
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    1) can be factored more the expression (x^4-x^4) is also a difference of squares 2) seems a bit off but yes use difference of squares 3) agreed used difference of cubes 4) use difference of squares

  5. anonymous
    • one year ago
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    Can you do solve no. 4??

  6. freckles
    • one year ago
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    \[u^2-v^2=(u-v)(u+v) \text{ is how to factored difference of squares } \\ u^3+v^3=(u+v)(u^2-uv+v^2) \text{ is how to factored sum of cubes } \\ u^3-v^3=(u-v)(u^2+uv+v^2) \text{ is how to factored difference of cubes }\]

  7. freckles
    • one year ago
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    \[\frac{a^2}{b^2}x^2-v^2 \\ \text{ where } v=u^2 \\ \text{ first all notice this can be written as } \\ (\frac{a}{b}x)^2-v^2 \\ \text{ this is a difference of squares }\]

  8. freckles
    • one year ago
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    example: \[\frac{16}{25}x^2-u^4 \\ \frac{4^2}{5^2}x^2-(u^2)^2 \\ (\frac{4}{5}x)^2-(u^2)^2 \\ (\frac{4}{5}x-u^2)(\frac{4}{5}x+u^2)\]

  9. freckles
    • one year ago
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    \[\frac{4}{9}x^2-y^4 \\ \frac{2^2}{3^2}x^2-(y^2)^2 \\ (\frac{2}{3}x)^2-(y^2)^2 \] can you try the factoring part?

  10. freckles
    • one year ago
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    it is just pluggin into a formula really

  11. anonymous
    • one year ago
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    Okay, i'll try that...

  12. anonymous
    • one year ago
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    Thank you so muuuch!!

  13. anonymous
    • one year ago
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    @freckles

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