Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$ 
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$

Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ and $\sqrt{x-1}\ge0.$ (I am asking if we can use in the inequalities after $x\ge1$ that $x$ is actually greater than or equal to $1$.)
After that we have $A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}.$ What from here? Thank you in advance!

The inequality $x>=1$ covers it all. You can simplify further by noting that
$\frac{1}{x+\sqrt{x-1}}=\frac{x-\sqrt{x-1}}{x^2-x+1}$
