Answer
$\dfrac{(x+3)(x-2)(x-5)}{(x-3)(x-1)(x+5)}$
Work Step by Step
By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have
$$
\frac{x^2+x-6}{x^2+4x-5}\frac{x^2-25}{x^2+2x-15}=\frac{(x+3)(x-2)}{(x+5)(x-1)}\frac{(x-5)(x+5)}{(x-3)(x+5)}\\
=\frac{(x+3)(x-2)(x-5)}{(x-3)(x-1)(x+5)}
.$$
