Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$

It's $$x^2(x-4)-4(x-4)=0$$ or $$(x-4)(x^2-4)=0.$$ Can you end it now?

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

We have that

$$x^3-4x^2-4x+16=x(x^2-4x+4)-8x+16=x(x-2)^2-8(x-2)=$$

$$=(x-2)(x^2-2x-8)=(x-2)(x+2)(x-4)=0$$