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(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?
 one year ago
 one year 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?
 one year ago
 one year ago

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MaledantBest ResponseYou've already chosen the best response.0
For the first one, no, for the second one, yes
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Works for first and second, question is why?
 one year ago

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

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

estudierBest ResponseYou've already chosen the best response.1
124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81
 one year ago

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

JonaskBest 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 \[x^2xy+y^2=x^2zx+z^2\] \[x(yz)=y^2z^2\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[5(14)=1^24^2\] true
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[67(4126)=41^226^2\] IS THAT TRUE
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Yes, that's true, what's the general rule, though?
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
this should be true\[x=yz\]
 one year ago

estudierBest ResponseYou've already chosen the best response.1
(n^3 + a^3)/(n^3 +b^3) where a+b = n
 one year ago

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

estudierBest ResponseYou've already chosen the best response.1
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[x(yz)=y^2z^2=(yz)(y+z)\] \[x=y+z\] but \[y \neq z\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????
 one year ago

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

JonaskBest ResponseYou've already chosen the best response.1
right,equal yes and without cubes
 one year ago
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