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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?
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?

This Question is Closed

Maledant
 3 years ago
Best ResponseYou've already chosen the best response.0For the first one, no, for the second one, yes

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1Works for first and second, question is why?

Maledant
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, let me correct that. Was wrong about my answer. forgot that 1 does not change even if cubed.

Maledant
 3 years ago
Best ResponseYou've already chosen the best response.0So correcting my self, yes, because \[1^3 = 1\]

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1this is working because of the factorisation of\[x^3+y^3\]

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ x^3+y^3 }{ x^3+z^3 }=\frac{ (x+y)(x^2xy+y^2) }{ (x+z)(x^2zx+ z^2)}\] which should equal\[\frac{ x+y }{ x+z }\] if and only if \[x^2xy+y^2=x^2zx+z^2\] \[x(yz)=y^2z^2\]

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1\[67(4126)=41^226^2\] IS THAT TRUE

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, that's true, what's the general rule, though?

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1this should be true\[x=yz\]

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1(n^3 + a^3)/(n^3 +b^3) where a+b = n

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1this should be true \[x=y+z\] the top one not true, a+b=n

estudier
 3 years ago
Best ResponseYou've already chosen the best response.11 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1\[x(yz)=y^2z^2=(yz)(y+z)\] \[x=y+z\] but \[y \neq z\]

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.11 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1K, you sub in what u r saying into the expansion of the cubes and then the denominator and numerator are equal, right?

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.1right,equal yes and without cubes
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