If $x^3 – 4x^2 + 19 = 6 (x – 1)$, then what is the value of $x^2 + (\frac{1}{x– 4})$.

Given: Expression  $x^3 – 4x^2 + 19 = 6 (x – 1)$.

To do: To the value of $x^2 + (\frac{1}{x} – 4)$.

Let's solve this:

$x^3-4x^2+19=6(x-1)\ \  ……( i)$

So we have to find the value of $x^2+\frac{1}{( x-4)}$

From equation $( i)$, we can say,

$x^3-4x^2+19=6( x-1)$

$\Rightarrow x^2(x-4)=6x-6-19$

$\Rightarrow x^2 (x-4)=6(x-4)-1$

$\Rightarrow x^2=\frac{6( x-4)-1}{( x-4)}$                  [dividing both the side by $( x-4)$]

$\Rightarrow x^2=6-\frac{1}{x-4}$

$\Rightarrow x^2+\frac{1}{( x-4)}=6$