Question

$( x^{2} -4) :\frac{3x^2+6x}{x^3} * \frac{12}{4-4x+x^2}$

Answer 1

$(x^{2}-4):\frac{3x^2+6x}{x^3}*\frac{12}{4-4x+x^2}=(x-2)(x+2)*\frac{x^3}{3x(x+2)}*\frac{12}{(x-2)^2}=\frac{12x^3}{3x(x-2)}=\frac{12x^2}{3(x-2)}$