Thực hiện các phép tính sau: $(\sqrt[3]{4}+1)^3$ $-$ $(\sqrt[3]{4}-1)^3$ $(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})$$(\sqrt[3]{3}+\sqrt[3]{2})$

Đáp án+Giải thích các bước giải:

$(\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3\\ =(\sqrt[3]{4})^3+3.(\sqrt[3]{4})^2.1+3.\sqrt[3]{4}.1^2+1^3-[(\sqrt[3]{4})^3-3.(\sqrt[3]{4})^2.1+3.\sqrt[3]{4}.1^2-1^3]\\ =4+3.(\sqrt[3]{4})^2+3\sqrt[3]{4}+1-[4-3.(\sqrt[3]{4})^2+3\sqrt[3]{4}-1]\\ =4+3.(\sqrt[3]{4})^2+3\sqrt[3]{4}+1-4+3.(\sqrt[3]{4})^2-3\sqrt[3]{4}+1\\ =2+6.(\sqrt[3]{4})^2\\ =2+6\sqrt[3]{4^2}\\ =2+6\sqrt[3]{16}\\ =2+6\sqrt[3]{8}.\sqrt[3]{2}\\ =2+6.2.\sqrt[3]{2}\\ =2+12\sqrt[3]{2}\\ (\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})\\ =(\sqrt[3]{3^2}-\sqrt[3]{3.2}+\sqrt[3]{2^2})(\sqrt[3]{3}+\sqrt[3]{2})\\ =[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\\ =(\sqrt[3]{3})^3+(\sqrt[3]{2})^3\\ =3+2\\ =5.$