Dạng 2: Thực hiện phép tính

• Phương pháp giải

Áp dụng công thức: \[\sqrt[3]{{{a}^{3}}}=a;{{\left( \sqrt[3]{a} \right)}^{3}}=a.\] Áp dụng: Các hằng đẳng thức: \[{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}};\] \[{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}};\] \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right);\] \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right).\]

Ví dụ. Thực hiện các phép tính sau a) \[\sqrt[3]{\left( \sqrt{2}+1 \right)\left( 3+2\sqrt{2} \right)}.\] b) \[\left( \sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4} \right)\left( \sqrt[3]{3}+\sqrt[3]{2} \right).\] Hướng dẫn giải a) \[\sqrt[3]{\left( \sqrt{2}+1 \right)\left( 3+2\sqrt{2} \right)}\] = \[\sqrt[3]{\left( \sqrt{2}+1 \right)\left( 2+2\sqrt{2}+1 \right)}\] \[=\sqrt[3]{\left( \sqrt{2}+1 \right){{\left( \sqrt{2}+1 \right)}^{2}}}\] \[=\sqrt[3]{{{\left( \sqrt{2}+1 \right)}^{3}}}\] \[=\sqrt{2}+1\] b) \[\left( \sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4} \right)\left( \sqrt[3]{3}+\sqrt[3]{2} \right)\] \[=\left( \sqrt[3]{3}+\sqrt[3]{2} \right)\left[ {{\left( \sqrt[3]{3} \right)}^{2}}-\sqrt[3]{3}.\sqrt[3]{2}+{{\left( \sqrt[3]{2} \right)}^{2}} \right]\] \[=3+2=5\]

• Ví dụ mẫu

Ví dụ 1.

Thực hiện các phép tính sau

a) \[\sqrt[3]{\left( 4-2\sqrt{3} \right)\left( \sqrt{3}-1 \right)}\]. b) \[\sqrt[3]{-64}-\sqrt[3]{125}+\sqrt[3]{216}\].

c) \[{{\left( \sqrt[3]{4}+1 \right)}^{3}}-{{\left( \sqrt[3]{4}-1 \right)}^{3}}\].

Hướng dẫn giải

a) \[\sqrt[3]{\left( 4-2\sqrt{3} \right)\left( \sqrt{3}-1 \right)}=\sqrt[3]{\left( 3-2\sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}\]

\[=\sqrt[3]{{{\left( \sqrt{3}-1 \right)}^{2}}\left( \sqrt{3}-1 \right)}=\sqrt[3]{{{\left( \sqrt{3}-1 \right)}^{3}}}\]

\[=\sqrt{3}-1.\]

b) \[\sqrt[3]{-64}-\sqrt[3]{125}+\sqrt[3]{216}=\sqrt[3]{{{\left( -4 \right)}^{3}}}-\sqrt[3]{{{5}^{3}}}+\sqrt[3]{{{6}^{3}}}=-4-5+6=-3\]

c) \[{{\left( \sqrt[3]{4}+1 \right)}^{3}}-{{\left( \sqrt[3]{4}-1 \right)}^{3}}\]

\[=\left[ \left( \sqrt[3]{4}+1 \right)-\left( \sqrt[3]{4}-1 \right) \right].\left[ {{\left( \sqrt[3]{4}+1 \right)}^{2}}+\left( \sqrt[3]{4}+1 \right).\left( \sqrt[3]{4}-1 \right)+{{\left( \sqrt[3]{4}-1 \right)}^{2}} \right]\]

\[=2.\left[ {{\left( \sqrt[3]{4} \right)}^{2}}+2\sqrt[3]{4}+1+{{\left( \sqrt[3]{4} \right)}^{2}}-1+{{\left( \sqrt[3]{4} \right)}^{2}}-2\sqrt[3]{4}+1 \right]\]

\[=2.\left( 3\sqrt[3]{16}+1 \right)\]

\[=12\sqrt[3]{2}+2\]

Ví dụ 2.

Tính \[A=\sqrt[3]{17\sqrt{5}+38}-\sqrt[3]{17\sqrt{5}-38}\]

Hướng dẫn giải

Nhận xét: \[\sqrt[3]{\left( 17\sqrt{5}+38 \right)}.\sqrt[3]{\left( 17\sqrt{5}-38 \right)}=1.\]

Đặt \[a=\sqrt[3]{17\sqrt{5}+38};b=\sqrt[3]{17\sqrt{5}-38}\], ta có $\left\{ \begin{array} & {{a}^{3}}-{{b}^{3}}=76 \\ a.b=1 \\ \end{array} \right.$

Xét \[{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)\]

\[\Rightarrow {{A}^{3}}=76-3A\Leftrightarrow {{A}^{3}}+3A-76=0\Leftrightarrow \left( A-4 \right)({{A}^{2}}+4A+19)=0\]

\[\Leftrightarrow (A-4)\left[ {{\left( A+2 \right)}^{2}}+15 \right]=0\]

\[\begin{array} & \Rightarrow A-4=0 \\ \Rightarrow A=4 \\ \end{array}\]

Vậy \[A=\sqrt[3]{17\sqrt{5}+38}-\sqrt[3]{17\sqrt{5}-38}\]=4.