## anonymous 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?

1. anonymous

yes

2. anonymous

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

3. anonymous

Works for first and second, question is why?

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

$\frac{ x^3+y^3 }{ x^3+z^3 }=\frac{ (x+y)(x^2-xy+y^2) }{ (x+z)(x^2-zx+ z^2)}$ 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. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

this should be true$x=y-z$

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. anonymous

right,equal yes and without cubes

20. anonymous

Cool.