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GoldRush18

  • 2 years ago

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

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  1. GoldRush18
    • 2 years ago
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    this is a Hard one for me

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

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

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

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

  6. experimentX
    • 2 years ago
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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!!!

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

  8. satellite73
    • 2 years ago
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    you got this?

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

  10. satellite73
    • 2 years ago
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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

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

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

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

  14. GoldRush18
    • 2 years ago
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    oh ok

  15. satellite73
    • 2 years ago
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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?

  16. satellite73
    • 2 years ago
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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

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

  18. satellite73
    • 2 years ago
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    yes

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

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

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

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

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

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

  25. satellite73
    • 2 years ago
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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

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

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

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

  29. GoldRush18
    • 2 years ago
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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

  30. satellite73
    • 2 years ago
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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

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

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

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

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

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

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

  37. experimentX
    • 2 years ago
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    yw!!

  38. satellite73
    • 2 years ago
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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

  39. satellite73
    • 2 years ago
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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

  40. satellite73
    • 2 years ago
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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

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

  42. satellite73
    • 2 years ago
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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

  43. experimentX
    • 2 years ago
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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!!

  44. satellite73
    • 2 years ago
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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

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

  46. satellite73
    • 2 years ago
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    nvm

  47. experimentX
    • 2 years ago
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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!!

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

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