anonymous
  • anonymous
(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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
yes
anonymous
  • anonymous
For the first one, no, for the second one, yes
anonymous
  • anonymous
Works for first and second, question is why?

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anonymous
  • anonymous
Oh, let me correct that. Was wrong about my answer. forgot that 1 does not change even if cubed.
anonymous
  • anonymous
So correcting my self, yes, because \[1^3 = 1\]
anonymous
  • anonymous
124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81
anonymous
  • anonymous
this is working because of the factorisation of\[x^3+y^3\]
anonymous
  • anonymous
\[\frac{ x^3+y^3 }{ x^3+z^3 }=\frac{ (x+y)(x^2-xy+y^2) }{ (x+z)(x^2-zx+ z^2)}\] which should equal\[\frac{ x+y }{ x+z }\] if and only if \[x^2-xy+y^2=x^2-zx+z^2\] \[x(y-z)=y^2-z^2\]
anonymous
  • anonymous
\[5(1-4)=1^2-4^2\] true
anonymous
  • anonymous
\[67(41-26)=41^2-26^2\] IS THAT TRUE
anonymous
  • anonymous
Yes, that's true, what's the general rule, though?
anonymous
  • anonymous
this should be true\[x=y-z\]
anonymous
  • anonymous
(n^3 + a^3)/(n^3 +b^3) where a+b = n
anonymous
  • anonymous
this should be true \[x=y+z\] the top one not true, a+b=n
anonymous
  • anonymous
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....
anonymous
  • anonymous
\[x(y-z)=y^2-z^2=(y-z)(y+z)\] \[x=y+z\] but \[y \neq z\]
anonymous
  • anonymous
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????
anonymous
  • anonymous
K, you sub in what u r saying into the expansion of the cubes and then the denominator and numerator are equal, right?
anonymous
  • anonymous
right,equal yes and without cubes
anonymous
  • anonymous
Cool.

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