estudier
(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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queenchatter
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yes
Maledant
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For the first one, no, for the second one, yes
estudier
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Works for first and second, question is why?
Maledant
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Oh, let me correct that. Was wrong about my answer. forgot that 1 does not change even if cubed.
Maledant
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So correcting my self, yes, because \[1^3 = 1\]
estudier
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124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81
Jonask
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this is working because of the factorisation of\[x^3+y^3\]
Jonask
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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
\[x^2-xy+y^2=x^2-zx+z^2\]
\[x(y-z)=y^2-z^2\]
Jonask
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\[5(1-4)=1^2-4^2\] true
Jonask
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\[67(41-26)=41^2-26^2\] IS THAT TRUE
estudier
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Yes, that's true, what's the general rule, though?
Jonask
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this should be true\[x=y-z\]
estudier
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(n^3 + a^3)/(n^3 +b^3) where a+b = n
Jonask
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this should be true \[x=y+z\]
the top one not true,
a+b=n
estudier
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1 + 4 = 5
41 +26 =67
43 +81 = 124
Still haven't (quite) proved it though....
Jonask
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\[x(y-z)=y^2-z^2=(y-z)(y+z)\]
\[x=y+z\]
but
\[y \neq z\]
Jonask
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1 + 4 = 5
41 +26 =67
43 +81 = 124
Still haven't (quite) proved it though....
????
estudier
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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?
Jonask
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right,equal yes and without cubes
estudier
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Cool.