Question 126328
(x^2-4/x^2-x-6) / (x^2-x-2/x^2-8x+15) * (x^2-3x-10/x^2+3x+2) 

[(x-2)(x+2)/(x-3)(x+2)] / [(x-2)(x+1)/(x-3)(x-5)] * [(x-5)(x+2)/(x+2)(x+1)]

Cancel where you can to get:

[ 1 / (x-3)] / [(x+1)/(x-3)(x-5)] * [(x-5)/(x+1)]

Invert the denominator to get:

[ 1 / (x-3)] * [(x-3)(x-5)/(x+1)] * [(x-5)/(x+1)]

Cancel again where you can to get:

[ 1 / 1] * [(x-5)/(x+1)] * [(x-5)/(x+1)]

= [(x-5)/(x+1)]^2

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Cheers,
Stan H.