Simplify, $\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$ where $x > 3$.

Question

Simplify, $$\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}$$ where $x > 3$.

What I Tried: I tried to rationalize the denominator by multiplying with $x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9})$ to get :- $$\Bigg(\frac{x^2 - 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 + (x - 1)\sqrt{(x^2 - 9)}} * \frac{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}{x^2 + 4x + 3 - (x - 1)\sqrt{(x^2 - 9)}}\Bigg)$$ $$\rightarrow \frac{2(x - 1)(x^2 + 4x\sqrt{x^2 - 9}x - 9)}{(x^2 + 4x + 3)^2 - (x - 1)^2(x + 3)(x - 3)}$$

I was only able to get to this, and I am stuck on how to move next. Can anyone help me?

Edit: The answer that is given in my side is $\sqrt{\frac{x^2 - 9}{x + 3}}$ , so for now none of the answers match with it.

Answer 1

$$\frac{(x-3)(x-1) +(x-1)\sqrt{(x-3)(x+3)}}{(x+3)(x+1) +(x-1)\sqrt{(x-3)(x+3)}}\\ = \frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}} \frac{\sqrt{x-3} +\sqrt{x+3}}{(x+1)\sqrt{x+3}+(x-1)\sqrt{x-3}}$$

Multiply top and bottom by $\sqrt{x+3}-\sqrt{x-3}$.

$$= \frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}}\frac{6}{8x-2\sqrt{x^2-9}} \\ = 3\frac{(x-1)\sqrt{x-3}}{\sqrt{x+3}} \frac{4x+\sqrt{x^2-9}}{15x^2+9} \\ = \frac{(x-1)\sqrt{x^2-9}}{x+3} \frac{4x+\sqrt{x^2-9}}{5x^2+3} \\ =\frac{(x-1)(x^2-9 +4x\sqrt{x^2-9})}{(5x^2+3)(x+3) } \\ =\frac{(x-1)(x-3)}{5x^2+3} +\frac{4x(x-1)}{(5x^2+3)(x+3)} \sqrt{x^2-9}$$

I doubt this can be simplified any further.

Answer 2

$(x^2+4x+3)^2 - (x-1)^2(x+3)(x-3) = (x+1)^2(x+3)^2 - (x-1)^2(x+3)(x-3) = (x+3)(10x^2 + 6)$

And $x^2 + 4x^2\sqrt{x^2-9} -9 = \sqrt{x^2-9}(\sqrt{x^2-9}+4x^2)$.

So you get $$\frac{\sqrt{x^2-9}(\sqrt{x^2-9}+4x^2)}{(x+3)(10x^2 + 6)} = \frac{\sqrt{x-3}(\sqrt{x^2-9}+4x^2)}{\sqrt{x+3}(10x^2 + 6)}$$

Or $$\frac{(x-3)(x+3)+4x^2\sqrt{x^2-9}}{(x+3)(10x^2 + 6)} = \frac{x-3}{10x^2+6} + \frac{4x^2\sqrt{x^2-9}}{(x+3)(10x^2 + 6)}$$