Find the approximate square root of $(a^4-a^{-4})/(a^2-a^{-2})$ Find the approximate square root of
$$\dfrac{\left( \dfrac{12}{5}\right)^4 - \left( \dfrac{5}{12}\right)^4  }{\left( \dfrac{12}{5}\right)^2 - \left( \dfrac{5}{12}\right)^2}$$
I tried using the formula for $(a^4-b^4)$ and $(a^2-b^2)$. Then cancelled the common $(a-b)$, substituted the values and simplified. Didn't work. 
Answer is 13 
Can someone tell me how to solve it with steps of possible? 
It will be a great help 
 A: Let $x= \frac{12}{5}$ to obtain
\begin{align}
f(x) &= \frac{x^{4} + \frac{1}{x^{4}}}{x^{2} + \frac{1}{x^{2}}} = \frac{x^{8}-1}{x^{2} \, (x^{4}-1)} = \frac{x^{4} + 1}{x^{2}} = x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} - 2 
\end{align}
Taking the square root of both sides leads to
\begin{align}
\sqrt{f(x)} = \sqrt{x^{2}+ \frac{1}{x^{2}}} = x \, \sqrt{1 + \frac{1}{x^{4}}} = x \, \left( 1 + \frac{1}{2 \, x^{4}} - \frac{1}{8 \, x^{8}} + \mathcal{O}\left(\frac{1}{x^{10}}\right) \right) \approx x
\end{align}
A: The fraction itself has a value of around $6$, so the square root of the fraction will be somewhere between $2$ and $3$, and not $13$. How did you get $13$?
The most I was able to simplify the value of the whole expression is that it is equal to $$13^2\left(\frac1{5^2} + \frac1{12^2}\right) - 2$$
Which, I guess, you could say is close to $$\frac{13^2}{5^2} - 2 = \frac{12^2-5^2}{5^2}$$ which is "close" to $\frac{12^2}{5^2}$ and the root of this is $\frac{12}5=2.4$ Which is actually pretty close to the actual answer!
A: First, this is straight equal to $\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2}$ through a simple cancellation.  Now notice that 
$$\left(\frac{12}{5}\right)^{2} + \left(\frac{5}{12}\right)^{2} =\left(\frac{12}{5}\right)^{2} + \left(\frac{12}{5}\right)^{-2} = \left(\frac{5}{12}\right)^{2} \left( \left(\frac{12}{5}\right)^{4} +1 \right)$$
Now take 
$$\sqrt{\left(\frac{5}{12}\right)^{2} \left( \left(\frac{12}{5}\right)^{4} +1 \right)} = \frac{5}{12}\sqrt{\left(\frac{12}{5}\right)^{4} +1 } \approx \frac{5}{12} \left(\frac{12}{5}\right)^{2} = \frac{12}{5} = 2.4$$
It's no problem to drop the $1$, because $1$ is very small in relation to $\left(\frac{12}{5}\right)^{4}$.
A: It can't be 13, in any case !
After cancelling out common terms, it becomes ${\left( \dfrac{12}{5}\right)^2 + \left( \dfrac{5}{12}\right)^2}$
Ignoring the 2nd term and taking the square root, ans = $\approx2.4$
See wolframalpha for a figure of 2.4359 to 4 dp
NOTE:
For a GRE test, you need a quick and dirty method to approximate!
