I been stuck on this for half an hour now and I need some help, I need to simplify the expression: $$\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}}$$
$\endgroup$ 52 Answers
$\begingroup$$$\require{cancel}\begin{align}\left[\left(\frac{a\sqrt{2}}{\left(1+a^{2}\right)^{-1}}\right) - \left(\frac{2\sqrt{2}}{a^{-1}}\right)\right] \cdot\frac{a^{-3}}{1-a^{-2}} &= \left[\sqrt{2}a\left(1+a^2\right)-2\sqrt{2}a\right]\frac{1}{a^3\left(1-a^{-2}\right)} \\ &= \frac{\left[\sqrt{2}\color{red}{\cancel a}\left(1+a^2\right)-2\sqrt{2}\color{red}{\cancel a}\right]}{\color{red}{\cancel a}\left(a^2-1\right)} \\ &= \frac{\left[\sqrt{2}\left(1+a^2\right)-2\sqrt{2}\right]}{\left(a^2-1\right)} \\ &= \frac{-\sqrt{2}+\sqrt{2}a^2}{\left(a^2-1\right)} \\ &= \frac{\sqrt{2}\color{red}{\cancel {\left(a^2-1\right)}}}{\color{red}{\cancel {\left(a^2-1\right)}}} \\ &= \sqrt{2}\end{align}$$
$\endgroup$ 1 $\begingroup$$$(a\sqrt{2}(1+a^2)-2\sqrt{2}a)\frac{1}{(a^2-1)a} =\frac{\sqrt{2}(a^2-1)}{(a^2-1)}=\sqrt{2}$$
EDIT $\frac{a^{-3}}{(1-a^{-2})}=\frac{\frac{1}{a^3}}{(1-\frac{1}{a^2})}=\frac{a^2}{a^3(a^2-1)}=\frac{1}{a(a^2-1)}$
$\endgroup$ 1
