anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Thank you for your answering and... yes it is divided by 9 in no. 4
anonymous
  • anonymous
is it really minus in squares on no. 1??? Im confused

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freckles
  • freckles
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
anonymous
  • anonymous
Can you do solve no. 4??
freckles
  • freckles
\[u^2-v^2=(u-v)(u+v) \text{ is how to factored difference of squares } \\ u^3+v^3=(u+v)(u^2-uv+v^2) \text{ is how to factored sum of cubes } \\ u^3-v^3=(u-v)(u^2+uv+v^2) \text{ is how to factored difference of cubes }\]
freckles
  • freckles
\[\frac{a^2}{b^2}x^2-v^2 \\ \text{ where } v=u^2 \\ \text{ first all notice this can be written as } \\ (\frac{a}{b}x)^2-v^2 \\ \text{ this is a difference of squares }\]
freckles
  • freckles
example: \[\frac{16}{25}x^2-u^4 \\ \frac{4^2}{5^2}x^2-(u^2)^2 \\ (\frac{4}{5}x)^2-(u^2)^2 \\ (\frac{4}{5}x-u^2)(\frac{4}{5}x+u^2)\]
freckles
  • freckles
\[\frac{4}{9}x^2-y^4 \\ \frac{2^2}{3^2}x^2-(y^2)^2 \\ (\frac{2}{3}x)^2-(y^2)^2 \] can you try the factoring part?
freckles
  • freckles
it is just pluggin into a formula really
anonymous
  • anonymous
Okay, i'll try that...
anonymous
  • anonymous
Thank you so muuuch!!
anonymous
  • anonymous
@freckles

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