Question

a ) \(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}\)

Answer 1

\(\frac{x+3}{x^2-4}.\frac{8-12x+6x^2-x^3}{9x+27}\)

\(=\frac{\left(x+3\right)\left(2-x\right)^3}{\left(x+2\right)\left(x-2\right).9\left(x+3\right)}\)

\(=-\frac{\left(x-2\right)^2}{9\left(x+2\right)}\)

Answer 2

\(\frac{x+3}{x^2-4}\cdot\frac{8-12x+6x^2-x^3}{9x+27}\)

\(=\frac{x+3}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(2-x\right)^3}{9\left(x+3\right)}\)

\(=-\frac{\left(2-x\right)^2}{9\left(x+2\right)}\)