# Simplify the given expression

I been stuck on this for half an hour now and I need some help, I need to simplify the expression: $$\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}}$$

• i can not read this Commented Nov 9, 2015 at 17:05
• It's not just about saying that you've done some efforts to solve to problem,it's about to prove it. Commented Nov 9, 2015 at 17:11
• @Nameless I cant prove because I couldn't came up with a sensible answer yet
– BsD
Commented Nov 9, 2015 at 17:19
• You can atleast show us where you got stuck. Anyways I have posted an answer. Commented Nov 9, 2015 at 17:23
• There's no need to come up with a sensible answer,it's just about to share your thoughts on the problem,to tell what are the crucial points which you find hard to tackle and ,if possible, to tell what you think the answer would be etc...so that you can learn the most. Commented Nov 9, 2015 at 17:23

\require{cancel}\begin{align}\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}} &= \left[\sqrt{2}a\left(1+a^2\right)-2\sqrt{2}a\right]\frac{1}{a^3\left(1-a^{-2}\right)} \\ &= \frac{\left[\sqrt{2}\color{red}{\cancel a}\left(1+a^2\right)-2\sqrt{2}\color{red}{\cancel a}\right]}{\color{red}{\cancel a}\left(a^2-1\right)} \\ &= \frac{\left[\sqrt{2}\left(1+a^2\right)-2\sqrt{2}\right]}{\left(a^2-1\right)} \\ &= \frac{-\sqrt{2}+\sqrt{2}a^2}{\left(a^2-1\right)} \\ &= \frac{\sqrt{2}\color{red}{\cancel {\left(a^2-1\right)}}}{\color{red}{\cancel {\left(a^2-1\right)}}} \\ &= \sqrt{2}\end{align}
$$(a\sqrt{2}(1+a^2)-2\sqrt{2}a)\frac{1}{(a^2-1)a} =\frac{\sqrt{2}(a^2-1)}{(a^2-1)}=\sqrt{2}$$
EDIT $\frac{a^{-3}}{(1-a^{-2})}=\frac{\frac{1}{a^3}}{(1-\frac{1}{a^2})}=\frac{a^2}{a^3(a^2-1)}=\frac{1}{a(a^2-1)}$