Question

выполните умножение
$(( {x}^{3} - 8) \div (x + 2)) \times (( {x}^{2} + 4x + 4) \div ( {x}^{2} + 2x + 4))$

Answer 1

$\frac{x^3-8}{x+2}\times\frac{x^2+4x+4}{x^2+2x+4}= \frac{x^3-2^3}{x+2}\times\frac{x^2+2*x*2+2^2}{x^2+2x+4}=\\\\ =\frac{(x-2)(x^2+x*2+2^2)}{x+2}\times\frac{(x+2)^2}{x^2+2x+4}=\\\\ =\frac{(x-2)(x^2+2x+4)*(x+2)^2}{(x+2)*(x^2+2x+4)}=\\\\ =\frac{(x-2)*(x+2)}{1}=x^2-2^2=\\\\ =x^2-4$