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estudier

  • 3 years ago

(5^3 +1) / (5^3 +4^3) = (5+1)/(5+4) (67^3 + 41^3) /(67^3 + 26^3) = (67 +41)/(67+26) So cancelling 3's works?

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  1. queenchatter
    • 3 years ago
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    yes

  2. Maledant
    • 3 years ago
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    For the first one, no, for the second one, yes

  3. estudier
    • 3 years ago
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    Works for first and second, question is why?

  4. Maledant
    • 3 years ago
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    Oh, let me correct that. Was wrong about my answer. forgot that 1 does not change even if cubed.

  5. Maledant
    • 3 years ago
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    So correcting my self, yes, because \[1^3 = 1\]

  6. estudier
    • 3 years ago
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    124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81

  7. Jonask
    • 3 years ago
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    this is working because of the factorisation of\[x^3+y^3\]

  8. Jonask
    • 3 years ago
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    \[\frac{ x^3+y^3 }{ x^3+z^3 }=\frac{ (x+y)(x^2-xy+y^2) }{ (x+z)(x^2-zx+ z^2)}\] which should equal\[\frac{ x+y }{ x+z }\] if and only if \[x^2-xy+y^2=x^2-zx+z^2\] \[x(y-z)=y^2-z^2\]

  9. Jonask
    • 3 years ago
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    \[5(1-4)=1^2-4^2\] true

  10. Jonask
    • 3 years ago
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    \[67(41-26)=41^2-26^2\] IS THAT TRUE

  11. estudier
    • 3 years ago
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    Yes, that's true, what's the general rule, though?

  12. Jonask
    • 3 years ago
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    this should be true\[x=y-z\]

  13. estudier
    • 3 years ago
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    (n^3 + a^3)/(n^3 +b^3) where a+b = n

  14. Jonask
    • 3 years ago
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    this should be true \[x=y+z\] the top one not true, a+b=n

  15. estudier
    • 3 years ago
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    1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....

  16. Jonask
    • 3 years ago
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    \[x(y-z)=y^2-z^2=(y-z)(y+z)\] \[x=y+z\] but \[y \neq z\]

  17. Jonask
    • 3 years ago
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    1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????

  18. estudier
    • 3 years ago
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    K, you sub in what u r saying into the expansion of the cubes and then the denominator and numerator are equal, right?

  19. Jonask
    • 3 years ago
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    right,equal yes and without cubes

  20. estudier
    • 3 years ago
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    Cool.

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