1. ## Simplify

Hi

I need to simplify the following, giving the result without fractional indices:

$\frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}}$

The solution: $(x+1)^2 \sqrt{x^2-1}$

I don't see how they obtain this solution, can somebody give me a few steps?

Thank you

2. Originally Posted by Coxx
Hi

I need to simplify the following, giving the result without fractional indices:

$\frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}}$

The solution: $(x+1)^2 \sqrt{x^2-1}$

I don't see how they obtain this solution, can somebody give me a few steps?

Thank you
simplify:
$\frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}}$
----------------------------------------------------------
remember:
$(a^2-b^2)=(a-b)(a+b)$
so,
$(x^2-1)^2= ((x-1)(x+1))^2=(x+1)^2 (x-1)^2$
-----------------------------------------------------------
$\frac{(x+1)^2 (x-1)^2 \sqrt{x+1}}{(x-1)^{3/2}}$

u can cancel (x-1)...

3. Hello, Coxx!

What a strange way to leave the answer . . .

Simplify: . $\frac{(x^2-1)^2 \sqrt{x+1}}{(x-1)^{3/2}}$

The solution: $(x+1)^2 \sqrt{x^2-1}$

We have: . $\frac{\bigg[(x-1)(x+1)\bigg]^2\sqrt{x+1}}{(x-1)^{\frac{3}{2}}}$ . $= \;\frac{(x-1)^2(x+1)^2(x+1)^{\frac{1}{2}}}{(x-1)^{\frac{3}{2}}}$

. . . . $=\; \frac{(x-1)^2}{(x-1)^{\frac{3}{2}}} \cdot\frac{(x+1)^2(x+1)^{\frac{1}{2}}}{1} \;=\;(x-1)^{\frac{1}{2}}(x+1)^{\frac{1}{2}}(x+1)^2$

. . . . $=\;\bigg[(x-1)(x+1)\bigg]^{\frac{1}{2}}(x+1)^2 \;=\;(x^2-1)^{\frac{1}{2}}(x+1)^2$

. . . . $= \;\sqrt{x^2-1}\,(x+1)^2$

4. hi,

So obvious many thanks

I just started to improve my maths skills so i'm really a newbie in this field!

greets