<s>[INST] 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.
 [/INST] 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
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(1)}\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(2)}\\=(m^2+2)(m^2-1)^{2-\frac 32}=\\=(m^2+2)(m^2-1)^{\frac 12}=\color{red}{(m^2+2)\sqrt{(m^2-1)}}$$
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