Question

\(\sqrt[3]{27+6\sqrt{21}}+\sqrt[3]{27-6\sqrt{21}}\)

Answer 1

Lời giải:

Đặt \(\sqrt[3]{27+6\sqrt{21}}=a; \sqrt[3]{27-6\sqrt{21}}=b\) thì ta cần tính tổng $A=a+b$.

Ta có:

$a^3+b^3=54$

\(ab=\sqrt[3]{(27+6\sqrt{21})(27-6\sqrt{21})}=-3\)

$A^3=(a+b)^3=a^3+b^3=3ab(a+b)=54+3(-3)A$

$\Leftrightarrow A^3=54-9A$

$\Leftrightarrow A^3+9A-54=0$

$\Leftrightarrow A^2(A-3)+3A(A-3)+18(A-3)=0$

$\Leftrightarrow (A^2+3A+18)(A-3)=0$

$\Leftrightarrow A-3=0$ (do $A^2+3A+18>0$)

$\Leftrightarrow A=3$