Answer:
[tex]\large\boxed{\dfrac{x+3}{x^2-x-12}\cdot\dfrac{x-4}{x^2-8x+16}=\dfrac{1}{x^2-8x+16}}[/tex]
Step-by-step explanation:
[tex]\dfrac{x+3}{x^2-x-12}\cdot\dfrac{x-4}{x^2-8x+16}=\dfrac{x+3}{x^2+3x-4x-12}\cdot\dfrac{x-4}{x^2-8x+16}\\\\=\dfrac{x+3}{x(x+3)-4(x+3)}\cdot\dfrac{x-4}{x^2-8x+16}=\dfrac{x+3}{(x+3)(x-4)}\cdot\dfrac{x-4}{x^2-8x+16}\\\\\text{cancel}\ (x+3)\ \text{and}\ (x-4)\\\\=\dfrac{1}{x^2-8x+16}[/tex]
