# A^3 + b^3 = c^3

1. Mar 8, 2008

### Loren Booda

Does the equation

a3 + b3 = c3

(where a, b and c are constants) have any general geometrical significance?

2. Mar 8, 2008

### John Creighto

Not sure. What about if we give a triangle depth and do something like the Pythagoras theorem but use area instead of length?

3. Mar 8, 2008

### tiny-tim

No way!

I don't think cubes ever arise in geometry.

Maybe in calculus, and in some branches of physics, but not in geometry.

4. Mar 8, 2008

### jostpuur

Are you interested about values $a,b,c\in\mathbb{N}$ or $\in\mathbb{R}$?

With natural numbers, there is an equivalent problem with small cubes of fixed size. Suppose you have c^3 small cubes, piled into a one bigger cube. Can you take these c^3 small cubes, and form two new big cubes that contain precisely all these small cubes? The Fermat's last theorem says that you cannot.

With real numbers, an equation x^3+y^3=1 describes a surface of a L^p ball with value p=3. Frankly, I still have not understood why real analysis deals so much with L^p spaces. I don't know what's their significance, yet

5. Mar 8, 2008

### HallsofIvy

Staff Emeritus
Well, so much for solid geometry!

6. Mar 8, 2008

7. Mar 8, 2008

### tiny-tim

oh, those cubes!
™​

8. Mar 8, 2008

Or spheres.

9. Mar 9, 2008

### tiny-tim

… or Russian dolls! …

10. Mar 19, 2008

### Bananaman3579

Ok tell me if theres a problem with this but I think this is the only way to solve it.
cube root 5^3+cube root 5^3= cube root 10^3
the cube root just cancels out the cubing so the a and b variables need to be the same and the c variable needs to be double the a or b.
I don't see what use this would have but thats the only way I've found to solve it.

11. Mar 19, 2008

### Bananaman3579

I'm not bananaman, but a friend and I think his reasoning is flawed somewhere but the equation seems reasonable. I still believe that the equation is universally unsolvable, basically forbidden. WC

12. Mar 19, 2008

### tiny-tim

… daylight come, and I wanna go home …

How many Bananamen are there?

…Where is Talleyman when we need him! …

13. Mar 19, 2008

### Alex48674

There is no solution. No 2 (natural number) cubes add up to another cube. In fact no solutions exist in x^n + y^n =z^n, where n is any number greater then 2, as stated by Fermat's last theorem (as proved by Andrew Wiles).

bananana you can't cancel out like this. One way to show this is wrong is

5^3 +5^3= 10^3

125+125=1000

250=1000

and this is clearly not true.

14. Mar 19, 2008

### Bananaman3579

I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.

15. Mar 19, 2008

### Bananaman3579

o and the "other" bananaman was just my friend using my account

16. Mar 19, 2008

### Alex48674

Ok sorry I didn't see that bit, but regardless it's the same.

lets say you can cancel out the cube then

a=x^1\2
b=y^1/2
c=z^1/2

then a^3 +b^3= c^3

according to you a+b=c

so lets say 1+1=2

then plug back into the formula you get 1+1=8 which is not true

the cube rrot is irrelevantly

17. Mar 19, 2008

### Bananaman3579

ok so I actually did this on a calculator, feel free to do so yourself, and the equation works. Now the problem i see with it is that it takes away from the "spirit" of the equation in the sense that it is just canceling out the main part of the equation and therefore basically destroying it. It reminds me of those verbal equations kids would say like, give me your favorite number then multiply by 2 add this and that, and the number would end up being the original through inverse operations. Basically i proved that there is a way to solve it but its pointless.

18. Mar 19, 2008

### Bananaman3579

ok you might also have missed the fact that the number that had a cube root was the variable, a= cube root of one b= cube root of 1 c= cube root of 2. then plug that into the equation.

19. Mar 19, 2008

### Bananaman3579

or when dividing by zero !!!!O KNOW BLACK HOLE!!!!!..........fixzzle

20. Mar 19, 2008

### Alex48674

Sorry I had to go and finish, I was going to say:

a=x^1\3
b=y^1/3
c=z^1/3 This takes care of you cube root problem

then a^3 +b^3= c^3

according to you a+b=c

so lets say 1+1=2

then plug back into the formula you get 1+1=8 which is not true

Then if you want to convert back to your cubed root variables you get

1 + 1= 512

My point is yes, for yours it works because basically what you are doing is picking numbers that will work for your method, but you can not apply this method to everything because it will not work for all cases in which a +b=c would also work for the cube of each.

Also the reason the cube root is irrelevantly is because every number has a cube root, so if we have the number a, the cube root could be x ( cube root a=x) and then simplify the problem so that x^3 + y^3=z^3, and I I showed this does not work. Look up fermats last theorem it disproves it all.