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 answer to your first question is yes, you can assume that $x\geq 1$ because of $\sqrt{x-1}$, as you said.
The second aim is also true: $x+\sqrt{x-1}\geq 0 $ because $x> 0$ and $\sqrt{x-1}\geq 0$.
So, the domain of your equation is $[1,+\infty)$.
Then, one thing you can do to simplify $A$ is to multiply and  divide by the conjugate of the divisor:
$$
A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}\dfrac{x-\sqrt{x-1}}{x-\sqrt{x-1}}
=\dfrac{(x+1)(x^2-x+1)(x+\sqrt{x-1})}{(x-\sqrt{x-1})(x+\sqrt{x-1})}
$$
Using the Difference of squares:
$$
A=\dfrac{(x+1)(x^2-x+1)(x-\sqrt{x-1})}{x^2-(\sqrt{x-1})^2)}
=\dfrac{(x+1)(x^2-x+1)(x-\sqrt{x-1})}{x^2-x+1}
=(x+1)(x-\sqrt{x-1})
$$
