<s>[INST] 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 
 [/INST] 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 $\left(\frac{12}{5}\right)^{4}$.
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