Đặt \(\sqrt[3]{b}=x\Rightarrow b=x^3\). Khi đó biểu thức B được biến đổi về dạng :
\(B=\left(\dfrac{x^3}{x^3+8}-\dfrac{4x^3}{\left(x+2\right)^3}\right).\left(\dfrac{1+\dfrac{2}{x}}{1-\dfrac{2}{x}}\right)^2-\dfrac{24}{x^3+8}\)
ĐK: \(x\ne0;x\ne\pm2\)
\(B=\left(\dfrac{x^3}{x^3+8}-\dfrac{4x^3}{\left(x+2\right)^3}\right).\left(\dfrac{x+2}{x-2}\right)^2-\dfrac{24}{x^3+8}\)
\(B=\dfrac{x^3\left(x+2\right)^2-4x^3\left(x^2-2x+4\right)}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)
\(=\dfrac{-3x^5+12x^4-12x^3}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)
\(=\dfrac{-3x^3\left(x-2\right)^2}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)
\(=\dfrac{-3x^3}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{24}{x^3+8}=\dfrac{-3x^3}{x^3+8}-\dfrac{24}{x^3+8}\)
\(=\dfrac{-3\left(x^3+8\right)}{x^3+8}=-3\)
