Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!

$$\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=\frac{(x-4)(x-2)(x+2)}{\sqrt{(x-4)(x-1)}}==\frac{\sqrt{x-4}(x-2)(x+2)}{\sqrt{x-1}}$$
The fraction will be $=0$ if the numerator is $0$ and the denominator is not $0$. That is true for $x=2,-2,4$.
