I'm not completely sure how to do this as I han't covered it :P
but here's some aid:
from that we get: (I'm not entirely sure)
1^x + 1^x + 6^x + 12^x = 32 (by adding/subtracting in the fractions)
and 1^anything is always 1
so
1 + 1 + 6^x + 12^x = 32
2 + 6^x + 12^x = 32
6^x + 12^x = 30
and that's all I've gotten up to since I haven't done this before
hope someoone else can help and please correct me if I've done anything wrong
I think you might be mistaken in that
$\displaystyle 6^{\frac{1}{2}}=\sqrt{6} \not= 6 + \frac{1}{2}$
The whole $\displaystyle \frac{1}{2}$ is an exponent, and they can't be added or subtracted easily.
By the way, looks like this is a really hard problem :/
Hello, perash!
I have an "eyeball" solution . . .
We have: .$\displaystyle \left(\sqrt{6}-\sqrt{5}\right)^x + \left(\sqrt{6} + \sqrt{5}\right)^x + \left(\sqrt{3} - \sqrt{2}\right)^x + \left(\sqrt{3} + \sqrt{2}\right)^x \;=\;32$$\displaystyle \left(6^{\frac{1}{2}} - 5^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}}
- 2^{\frac{1}{2}}\right)^x + \left(3^{\frac{1}{2}} + 2^{\frac{1}{2}}\right)^x + \left(6^{\frac{1}{2}} + 5^{\frac{1}{2}}\right)^x \;=\;32$
Since the left side equals an integer, $\displaystyle x$ must be even: .$\displaystyle x = 2n$
We have: .$\displaystyle \left(\sqrt{6}-\sqrt{5}\right)^{2n} + \left(\sqrt{6} + \sqrt{5}\right)^{2n} + \left(\sqrt{3} - \sqrt{2}\right)^{2n} + \left(\sqrt{3} + \sqrt{2}\right)^{2n} \;=\;32$
. . $\displaystyle \left[\left(\sqrt{6}-\sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{6} + \sqrt{5}\right)^2\right]^n + \left[\left(\sqrt{3} - \sqrt{2}\right)^2\right]^n + \left[\left(\sqrt{3} + \sqrt{2}\right)^2\right]^n \;=\;32$
. . $\displaystyle \left(11 - 2\sqrt{30}\right)^n + \left(11 + 2\sqrt{30}\right)^n + \left(5 - 2\sqrt{6}\right)^n + \left(5 + 2\sqrt{6}\right)^n \;=\;32$
And we see that $\displaystyle n = 1$ is a solution.
Therefore: .$\displaystyle x \:=\:2n \:=\:\boxed{2}$
In retrospect we should have expected that because:
$\displaystyle \sqrt{6} - \sqrt{5} = \frac{\sqrt{6} - \sqrt{5}}{1} \cdot \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}}$
$\displaystyle = \frac{6 - 5}{\sqrt{6} + \sqrt{5}} = \frac{1}{\sqrt{6} + \sqrt{5}}$
The same holds for the $\displaystyle \sqrt{3} - \sqrt{2}$ term so we have:
$\displaystyle (\sqrt{6} - \sqrt{5})^x + (\sqrt{6} + \sqrt{5})^x + (\sqrt{3} - \sqrt{2})^x + (\sqrt{3} + \sqrt{2})^x$
$\displaystyle = \frac{1}{(\sqrt{6} + \sqrt{5})^x} + (\sqrt{6} + \sqrt{5})^x + \frac{1}{(\sqrt{3} + \sqrt{2})^x} + (\sqrt{3} + \sqrt{2})^x$
So if we have a + x solution, we should also have the - x solution also.
-Dan