Giải câu 7 trang 37 sách phát triển năng lực toán 9 tập 1

Thay $B=\sqrt[3]{1007+\sqrt{1014048}}+\sqrt[3]{1007-\sqrt{1014048}}$ vào biểu thức $B^{3}-3B+2$ ta có:

$(\sqrt[3]{1007+\sqrt{1014048}}+\sqrt[3]{1007-\sqrt{1014048}})^{3}$

- $3.(\sqrt[3]{1007+\sqrt{1014048}}+\sqrt[3]{1007-\sqrt{1014048}})+2$

= $1007+\sqrt{1014048}+3.(\sqrt[3]{1007+\sqrt{1014048}})^{2}.\sqrt[3]{1007-\sqrt{1014048}}$

+ $3.\sqrt[3]{1007+\sqrt{1014048}}.(\sqrt[3]{1007-\sqrt{1014048}})^{2}$ +

+ $1007-\sqrt{1014048}-3.\sqrt[3]{1007+\sqrt{1014048}}-3.\sqrt[3]{1007-\sqrt{1014048}}+2$

= $2014+3.\sqrt[3]{1007+\sqrt{1014048}}.\sqrt[3]{(1007+\sqrt{1014048}).(1007-\sqrt{1014048})}$

+ $3.\sqrt[3]{1007-\sqrt{1014048}}.\sqrt[3]{(1007+\sqrt{1014048}).(1007-\sqrt{1014048})}$

- $3.\sqrt[3]{1007+\sqrt{1014048}}-3.\sqrt[3]{1007-\sqrt{1014048}}+2$

= $2016+3.\sqrt[3]{1007+\sqrt{1014048}}.\sqrt[3]{1014049-1014048}$

+ $3.\sqrt[3]{1007-\sqrt{1014048}}.\sqrt[3]{1014049-1014048}$ 

- $3.\sqrt[3]{1007+\sqrt{1014048}}-3.\sqrt[3]{1007-\sqrt{1014048}}$

= $2016+3.\sqrt[3]{1007+\sqrt{1014048}}+3.\sqrt[3]{1007-\sqrt{1014048}}$

- $3.\sqrt[3]{1007+\sqrt{1014048}}-3.\sqrt[3]{1007-\sqrt{1014048}}$

= 2016