Exponents and rationals $$
\large\sqrt{8+\sqrt{5}-\sqrt{6-2\sqrt{5}}}
$$
I am asked to write the following as a ratio of two integers (i.e.,) show that this value is rational
(Do not use calculator)
What I tried doing was putting everything to the exponent of $.5$ and then was unsure how to proceed 
 A: Hint: $6-2\sqrt{5}=(\sqrt{5}-1)^2$. 
A: Andre's hint is great, but more generally, if you need to show that a quantity containing a square root is rational, try completing the square on the quantity under the root. 
The most inside root is best to start with because it's easier to analyze and things might cancel.
For example, to do that on $6 - 2\sqrt{5}$, you square $\sqrt{5}$ to get $5$, so $6 - 2\sqrt{5} + 5 - 5 = 5 - 2\sqrt{5} + 1 = (\sqrt{5}-1)^2$.
A: Since you have already been given an answer, let me share this nice algorithm to denest radicals due to Bill Dubuque. 
Write $\alpha={a+b\sqrt c}$ and denote $\alpha'={a-b\sqrt c}$. We denote the product $\alpha\cdot \alpha'=a^2-cb^2$ by the symbol $|\alpha|$, and we call this the norm of $\alpha$. The number $\alpha'$ can be called its conjugate. By the trace of $\alpha$ we mean the sum $\alpha+\alpha'=2a$. We may denote this by ${\rm tr }\;\alpha$. Then, let $\beta=\alpha-\sqrt{|\alpha|}$. 
Bill D. tells us that $$\sqrt \alpha=\frac{\beta}{\sqrt{{\rm tr}\; {\beta}}}$$
In your case, ${6-2\sqrt 5}$. This as norm $36-4\cdot 5=16$, so the square of this is $4$. Now, we get $\alpha-|\alpha|=2-2\sqrt 5$. This has trace $=2\cdot 2=4$, whose square is $2$, so our final answer is that $$\sqrt{6-2\sqrt 5}=\frac{2-2\sqrt 5}{2}=1-\sqrt 5$$ Since I can't ask Bill, I'll take we must fix the sign of the root ourselves, so we get it is $\sqrt 5-1>0$. Then 
$$\sqrt{8+\sqrt{5}-\sqrt{6-2\sqrt{5}}}$$ becomes $$\sqrt {8 + \sqrt 5  - \left( {\sqrt 5  - 1} \right)}  = \sqrt {8 + 1}  = \sqrt 9  = 3$$
