<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] Answer :
$\frac{x^3-4x^2-4x+16}{\sqrt{x^2 - 5x+4}}= \frac{x(x^2 - 4)-4(x^2 - 4)}{\sqrt{x^2 - 5x+4}}$=$\frac{(x^2 - 4)(x-4)}{\sqrt{x^2 - 5x+4}}$
$\sqrt{x^2 - 5x+4} = 0 $ if $  x =(1, 4) $
Suppose $x≠(1,4)$
$(x^2 - 4)(x-4)$ =0
$\Rightarrow$ $ (x - 2)(x+2)(x-4)=0$
$\Rightarrow $ the solution is ($x=2$ or $ x=- 2)$
Because if $x =4  $the denominator equal $0$
So $x=4$ not a solution
Finally :
$S=(2,-2) $
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