Recent content by Bananaman3579

thanks that was pretty much exactly what i pictured in my head

wasnt there an error in Wiles work too?

*I would know if there is any actual study based on what I'm trying to explain.

Three Dimensional Angles?!? ok, this is a stretch. I have an idea. If there are angles, measured in units of 1/360th of a circle. what happens when you add the third dimension factor. I thought it would be somehow measured in units of like the area of a sphere divided by 360^2 maybe. The...

Yeah I might be in a little over my head here but all I was finding was a set of numbers that would make that equation true, NOT a solution that would work infinitly. basically what I'm saying is that you can take any two numbers, take th cube root of them and plug that into A and B. thus making...

Ok I think your still missing something. First things first can we at least agree there is no application for this no matter what, lol. Moving on to the utterly pointless. You switched your variables a bit. Looking at what you said in your most recent post x+y=z not a+b=c because then the...

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

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.

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...

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

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...

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

Ok tell me if there's 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...