<s>[INST] 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!
 [/INST] Use the fact that\begin{align}x^3-4x^2-4x+16=0&\iff x(x^2-4)-4(x^2-4)=0\\&\iff(x-4)(x^2-4)=0.\end{align}
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