Answer
$ \frac{x-2}{x+5}$.
Work Step by Step
First factor all numerators and denominators.
Factor $x^2+3x-10$.
Rewrite the middle term $3x$ as $5x-2x$.
$\Rightarrow x^2+5x-2x-10$
Group the terms.
$\Rightarrow (x^2+5x)+(-2x-10)$
Factor each group.
$\Rightarrow x(x+5)-2(x+5)$
Factor out $(x+5)$.
$\Rightarrow (x+5)(x-2)$
Factor $x^2+4x+3$.
Rewrite the middle term $4x$ as $3x+1x$.
$\Rightarrow x^2+3x+1x+3$
Group the terms.
$\Rightarrow (x^2+3x)+(1x+3)$
Factor each group.
$\Rightarrow x(x+3)+1(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x+1)$
Factor $x^2+x-6$.
Rewrite the middle term $x$ as $3x-2x$.
$\Rightarrow x^2+3x-2x-6$
Group the terms.
$\Rightarrow (x^2+3x)+(-2x-6)$
Factor each group.
$\Rightarrow x(x+3)-2(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x-2)$
Factor $x^2+10x+25$.
Use the special formula $(A+B)^2=A^2+2AB+B^2$.
$\Rightarrow (x+5)^2$
Back substitute into the given expression.
$\Rightarrow \frac{x^2+3x-10}{x^2+4x+3}\cdot \frac{x^2+x-6}{x^2+10x+25}\cdot \frac{x+1}{x-2}=\frac{(x+5)(x-2)}{(x+3)(x+1)}\cdot \frac{(x+3)(x-2)}{(x+5)(x+5)}\cdot \frac{x+1}{x-2}$
Cancel common factors.
$\Rightarrow \frac{x-2}{x+5}$.
