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.

• Just consider $(m-1)(m+1)=m^2-1$. Commented Jan 9, 2018 at 17:42
• It's false in general. $m\neq \pm 1$ Commented Jan 9, 2018 at 17:54

Write $(m-1)^2(m+1)^2$ as $(m^2-1)^2$.
Then you have $$\frac{(m^2-1)^2}{(m^2-1)^\frac{3}{2}}=\frac{(m^2-1)^\frac{4}{2}}{(m^2-1)^\frac{3}{2}}=(m^2-1)^{\frac{4}{2}-\frac{3}{2}}=(m^2-1)^\frac{1}{2}=\sqrt{m^2-1}$$
The $(m^2+2)$ part stays intact.

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

$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(1)}\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}\overbrace{=}^{(2)}\\=(m^2+2)(m^2-1)^{2-\frac 32}=\\=(m^2+2)(m^2-1)^{\frac 12}=\color{red}{(m^2+2)\sqrt{(m^2-1)}}$$

Simply note that

$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}=(m^2 + 2)(m^2-1)^{2-\frac32}=(m^2 + 2)(m^2-1)^{-\frac12}$$