Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expression.
please explain to me step by step. I'm newbie.

In order to obtain the second expression you have to know that:
$$a^2b^2=(ab)^2\implies(m-1)^2(m+1)^2=((m+1)(m+1))^2=(m^2-1)^2\tag{1}$$
$$\frac{a^x}{a^y}=a^{(x-y)}\tag{2}$$
Hence
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(1)}\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(2)}\\=(m^2+2)(m^2-1)^{2-\frac 32}=\\=(m^2+2)(m^2-1)^{\frac 12}=\color{red}{(m^2+2)\sqrt{(m^2-1)}}$$
