1. ## X^(1/3 -1/3)

What would happen if

X^(1/3 - 1/3)

I know that X^(1/3 - 2/3) = X^(-1/3)

What would happen then. It can't be 0/3.

2. Originally Posted by azurephantom
What would happen if

X^(1/3 - 1/3)

I know that X^(1/3 - 2/3) = X^(-1/3)

What would happen then. It can't be 0/3.
got news for you, it is ...

$\displaystyle x^{\frac{1}{3} - \frac{1}{3}} = x^0 = 1$

3. Originally Posted by azurephantom
What would happen if

X^(1/3 - 1/3)

I know that X^(1/3 - 2/3) = X^(-1/3)

What would happen then. It can't be 0/3.
Well, you probably shouldn't write 0/3's - but thats what it is, which is just 0.

4. Awesome I can use this to kill off some parts of a problem.

5. Originally Posted by skeeter
got news for you, it is ...

$\displaystyle x^{\frac{1}{3} - \frac{1}{3}} = x^0 = 1$
Yes!... of course!... but an interesting question: is that true for any real [or even complex...] $\displaystyle x$?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Originally Posted by chisigma
Yes!... of course!... but an interesting question: is that true for any real [or even complex...] $\displaystyle x$?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
not true for x = 0 ... what do you think about complex values of x?

7. The problem I intend to propose is the following: for real or complex $\displaystyle x$ and real or complex $\displaystyle a$ is...

$\displaystyle x^{a - a} = \frac {x^{a}}{x^{a}}$

... that means a number devided by himself. Well!... and whay is not...

$\displaystyle \frac {x^{a}}{x^{a}} = 1$

... for all $\displaystyle x$?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$