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?

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

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

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

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

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

11. estudier

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

this should be true$x=y-z$

13. estudier

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

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....

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

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?

right,equal yes and without cubes

20. estudier

Cool.