I been stuck on this for half an hour now and I need some help, I need to simplify the expression: $$\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}}$$
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$\begingroup$ i can not read this $\endgroup$– Dr. Sonnhard GraubnerCommented Nov 9, 2015 at 17:05
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$\begingroup$ It's not just about saying that you've done some efforts to solve to problem,it's about to prove it. $\endgroup$– NamelessCommented Nov 9, 2015 at 17:11
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$\begingroup$ @Nameless I cant prove because I couldn't came up with a sensible answer yet $\endgroup$– BsDCommented Nov 9, 2015 at 17:19
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$\begingroup$ You can atleast show us where you got stuck. Anyways I have posted an answer. $\endgroup$– Kushal BhuyanCommented Nov 9, 2015 at 17:23
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$\begingroup$ 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. $\endgroup$– NamelessCommented Nov 9, 2015 at 17:23
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2 Answers
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$$\require{cancel}\begin{align}\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}} &= \left[\sqrt{2}a\left(1+a^2\right)-2\sqrt{2}a\right]\frac{1}{a^3\left(1-a^{-2}\right)} \\ &= \frac{\left[\sqrt{2}\color{red}{\cancel a}\left(1+a^2\right)-2\sqrt{2}\color{red}{\cancel a}\right]}{\color{red}{\cancel a}\left(a^2-1\right)} \\ &= \frac{\left[\sqrt{2}\left(1+a^2\right)-2\sqrt{2}\right]}{\left(a^2-1\right)} \\ &= \frac{-\sqrt{2}+\sqrt{2}a^2}{\left(a^2-1\right)} \\ &= \frac{\sqrt{2}\color{red}{\cancel {\left(a^2-1\right)}}}{\color{red}{\cancel {\left(a^2-1\right)}}} \\ &= \sqrt{2}\end{align}$$
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$$(a\sqrt{2}(1+a^2)-2\sqrt{2}a)\frac{1}{(a^2-1)a} =\frac{\sqrt{2}(a^2-1)}{(a^2-1)}=\sqrt{2}$$
EDIT $\frac{a^{-3}}{(1-a^{-2})}=\frac{\frac{1}{a^3}}{(1-\frac{1}{a^2})}=\frac{a^2}{a^3(a^2-1)}=\frac{1}{a(a^2-1)}$
answered Nov 9, 2015 at 17:14
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$\begingroup$ Can you include the omitted algorithm because I cant understand how you got 1/(a^2-1)a as the second multiplier. $\endgroup$– BsDCommented Nov 9, 2015 at 17:27