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GoldRush18 Group Title

let x and y be positive real numbers such that x does not equal to y. i. simplify x^4-y^4/ x-y ii. hence, or otherwise, show that (y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3. iii. Deduce that (y+1)^4-y^4 < 4(y+1)^3

  • 2 years ago
  • 2 years ago

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  1. GoldRush18 Group Title
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    this is a Hard one for me

    • 2 years ago
  2. experimentX Group Title
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    ??? \[ x^4 - \frac{y^4}{x-y}\]

    • 2 years ago
  3. GoldRush18 Group Title
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    let me try to write that one better hold on

    • 2 years ago
  4. experimentX Group Title
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    \[ \frac{x^4 - y^4}{x-y}\]

    • 2 years ago
  5. GoldRush18 Group Title
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    \[x^4-y^4\div x-y\] Simplify

    • 2 years ago
  6. experimentX Group Title
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    (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) I think you can do the rest!!!

    • 2 years ago
  7. GoldRush18 Group Title
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    well this question has 3 parts to it

    • 2 years ago
  8. satellite73 Group Title
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    you got this?

    • 2 years ago
  9. GoldRush18 Group Title
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    nope i dont fully understand it

    • 2 years ago
  10. satellite73 Group Title
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    \[\frac{x^4 - y^4}{x-y}=\frac{(x+y)(x-y)(x^2+y^2)}{x-y}=(x+y)(x^2+y^2)\] is the first part

    • 2 years ago
  11. satellite73 Group Title
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    now replace \(x\) by \(y+1\)

    • 2 years ago
  12. experimentX Group Title
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    LOL ... i forgot the question!!

    • 2 years ago
  13. satellite73 Group Title
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    denominator is \(y+1-y=1\) numerator is \[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\]

    • 2 years ago
  14. GoldRush18 Group Title
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    oh ok

    • 2 years ago
  15. satellite73 Group Title
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    it is not clear to me what you wrote for the second part, but now algebra should give us what want. what was the second part supposed to be?

    • 2 years ago
  16. satellite73 Group Title
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    you wrote \[(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3.\] but the parentheses are messed up

    • 2 years ago
  17. GoldRush18 Group Title
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    so replace x=y+1 into the first question?

    • 2 years ago
  18. satellite73 Group Title
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    yes

    • 2 years ago
  19. GoldRush18 Group Title
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    thats what i see on the paper and i double checked

    • 2 years ago
  20. GoldRush18 Group Title
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    either way i think its correct nonetheless

    • 2 years ago
  21. satellite73 Group Title
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    \[(y+1)^4-y^4=(y+1+y)((y+1)^2+y^2)\] \[=((y+1)+y)((y+1)^2+y^2)\] \[=(y+1)^3+y(y+1)^2+y^2(y+1)+y^2\]

    • 2 years ago
  22. GoldRush18 Group Title
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    is this for the third part

    • 2 years ago
  23. satellite73 Group Title
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    i do not get what you want for the second part. there is a hanging plus sign and an extraneous )

    • 2 years ago
  24. GoldRush18 Group Title
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    i got the second part alright its the third part i don't get let me rewrite it here

    • 2 years ago
  25. satellite73 Group Title
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    @experimentX do you know what the second part is supposed to be? it makes no sense to me. there is a typo of some sort

    • 2 years ago
  26. GoldRush18 Group Title
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    deduce that \[(y+1)^4-y^4<4(y+1)^3\]

    • 2 years ago
  27. experimentX Group Title
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    sorry .. i was busy somewhere!! i'll see

    • 2 years ago
  28. satellite73 Group Title
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    so far i have this \[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\]

    • 2 years ago
  29. GoldRush18 Group Title
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    the second one to me is just simply saying the answer u got in first one just replace x=y+1 thats all

    • 2 years ago
  30. satellite73 Group Title
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    yes, but it has a specific form it wants you to put it in, and i don't know what that form is supposed to be

    • 2 years ago
  31. experimentX Group Title
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    y^3<(y+1)^3 y(y+1)^2<(y+1)^3 so on ..

    • 2 years ago
  32. experimentX Group Title
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    y^2(y+1) < (y+1)^3 so, \[ (y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < (y+1)^3 + (y+1)^3 + (y+1)^3 + (y+1)^3 = 4(y+1)^3\]

    • 2 years ago
  33. experimentX Group Title
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    looks like equation went off screen \[(y+1)^3+y(y+1)^2+y^2(y+1)+y^2 < 4(y+1)^3 \]

    • 2 years ago
  34. experimentX Group Title
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    so, \[ (y+1)^4 - y^4 < 4(y+1)^3\]

    • 2 years ago
  35. GoldRush18 Group Title
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    OK i get it now

    • 2 years ago
  36. GoldRush18 Group Title
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    thanks @satellite73 and @experimentX :)

    • 2 years ago
  37. experimentX Group Title
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    yw!!

    • 2 years ago
  38. satellite73 Group Title
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    maybe it is supposed to be this: \[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\] well i don't

    • 2 years ago
  39. satellite73 Group Title
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    @experimentX do you have any idea what the right side is supposed to be in equation 2? i mean i get the inequality, just not what the expression is supposed to be

    • 2 years ago
  40. satellite73 Group Title
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    this line \[(y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3\] has no meaning to me

    • 2 years ago
  41. experimentX Group Title
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    you mean equation 2 ??

    • 2 years ago
  42. satellite73 Group Title
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    i got this \[(y+1)^4-y^4=(y+1)^3+y(y+1)^2+y^2(y+1)+y^3\] , but it is not clear to me what we are supposed to turn it in to

    • 2 years ago
  43. experimentX Group Title
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    i think there a typho there!! a^n - b^n = (a - n) (a^(n-1) + a^(n-2)b + a^(n-3)b^2 + ... +b^(n-1)) this is the pattern!!

    • 2 years ago
  44. satellite73 Group Title
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    ok i got that. i won't worry about it i know there is a typo but i am not sure what it was supposed to be instead

    • 2 years ago
  45. GoldRush18 Group Title
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    maybe there was a typo because in maths alot of errors are made

    • 2 years ago
  46. satellite73 Group Title
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    nvm

    • 2 years ago
  47. experimentX Group Title
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    \( (y+1)^4-y^4=(y+1)^3+(y+1)^2+)y+1)y^2+y^3. \) ^ \( (y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3. \) ^ also, \( (y+1)^4-y^4=(y+1)^3+(y+1)^2+(y+1)y^2+y^3. \) ^ <---- y missing here .... i think ti should be what you got!!

    • 2 years ago
  48. satellite73 Group Title
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    got it. backwards parentheses and missing y confused the heck out of me

    • 2 years ago
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