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## 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

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1. queenchatter

yes

2. Maledant

For the first one, no, for the second one, yes

3. estudier

Works for first and second, question is why?

4. Maledant

Oh, let me correct that. Was wrong about my answer. forgot that 1 does not change even if cubed.

5. Maledant

So correcting my self, yes, because $1^3 = 1$

6. estudier

124^3 + 43^3/ 124^3 +81^3 = 124 +43/124 +81

7. Jonask

this is working because of the factorisation of$x^3+y^3$

8. Jonask

$\frac{ x^3+y^3 }{ x^3+z^3 }=\frac{ (x+y)(x^2-xy+y^2) }{ (x+z)(x^2-zx+ z^2)}$ which should equal$\frac{ x+y }{ x+z }$ if and only if $x^2-xy+y^2=x^2-zx+z^2$ $x(y-z)=y^2-z^2$

9. Jonask

$5(1-4)=1^2-4^2$ true

10. Jonask

$67(41-26)=41^2-26^2$ IS THAT TRUE

11. estudier

Yes, that's true, what's the general rule, though?

12. Jonask

this should be true$x=y-z$

13. estudier

(n^3 + a^3)/(n^3 +b^3) where a+b = n

14. Jonask

this should be true $x=y+z$ the top one not true, a+b=n

15. estudier

1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though....

16. Jonask

$x(y-z)=y^2-z^2=(y-z)(y+z)$ $x=y+z$ but $y \neq z$

17. Jonask

1 + 4 = 5 41 +26 =67 43 +81 = 124 Still haven't (quite) proved it though.... ????

18. estudier

K, you sub in what u r saying into the expansion of the cubes and then the denominator and numerator are equal, right?

19. Jonask

right,equal yes and without cubes

20. estudier

Cool.