Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
estudier
Group Title
(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?
 2 years ago
 2 years ago
estudier Group Title
(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?
 2 years ago
 2 years ago

This Question is Closed

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

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

Maledant Group TitleBest 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.
 2 years ago

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

estudier Group TitleBest ResponseYou've already chosen the best response.1
124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81
 2 years ago

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

Jonask Group TitleBest 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 \[x^2xy+y^2=x^2zx+z^2\] \[x(yz)=y^2z^2\]
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[5(14)=1^24^2\] true
 2 years ago

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

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

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

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

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

estudier Group TitleBest ResponseYou've already chosen the best response.1
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.1
1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????
 2 years ago

estudier Group TitleBest 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?
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
right,equal yes and without cubes
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.