Answer
$\frac{x+1}{x(x-1)}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{x^3-4x^2+x-4}{2x^3-8x^2+x-4}\cdot \frac{2x^3+2x^2+x+1}{x^4-x^3+x^2-x}$
Factor each numerator and denominator as shown below.
$\Rightarrow x^3-4x^2+x-4$
Group terms.
$\Rightarrow (x^3-4x^2)+(x-4)$
Factor each group.
$\Rightarrow x^2(x-4)+1(x-4)$
Factor out $(x-4)$.
$\Rightarrow (x-4)(x^2+1)$
$\Rightarrow 2x^3-8x^2+x-4$
Group terms.
$\Rightarrow (2x^3-8x^2)+(x-4)$
Factor each group.
$\Rightarrow 2x^2(x-4)+1(x-4)$
Factor out $(x-4)$.
$\Rightarrow (x-4)(2x^2+1)$
$\Rightarrow 2x^3+2x^2+x+1$
Group terms.
$\Rightarrow (2x^3+2x^2)+(x+1)$
Factor each group.
$\Rightarrow 2x^2(x+1)+1(x+1)$
Factor out $(x+1)$.
$\Rightarrow (x+1)(2x^2+1)$
$\Rightarrow x^4-x^3+x^2-x$
Factor out $x$ from each term.
$\Rightarrow x(x^3-x^2+x-1)$
Group terms.
$\Rightarrow x[(x^3-x^2)+(x-1)]$
Factor each group.
$\Rightarrow x[x^2(x-1)+1(x-1)]$
Factor out $(x-1)$.
$\Rightarrow x(x-1)(x^2+1)$
Substitute all factors into the given expression.
$\Rightarrow \frac{(x-4)(x^2+1)}{(x-4)(2x^2+1)}\cdot \frac{(x+1)(2x^2+1)}{x(x-1)(x^2+1)}$
Cancel common terms.
$\Rightarrow \frac{x+1}{x(x-1)}$.
