Sử dụng hằng đẳng thức $(a+b)^3 = (a^3 +b^3 ) + 3ab (a+b)$
Ta có $A=\sqrt[3]{6+\sqrt{\dfrac{847}{27}}}+\sqrt[3]{6+\sqrt{\dfrac{847}{27}}}$
$\Rightarrow A^3=\bigg ( \sqrt[3]{6+\sqrt{\dfrac{847}{27}}}+\sqrt[3]{6-\sqrt{\dfrac{847}{27}}} \bigg)^3$
$\Leftrightarrow A^3=12 +3\sqrt[3]{(6+\sqrt{\dfrac{847}{27}})(6-\sqrt{\dfrac{847}{27}})} \bigg [ \sqrt[3]{6+\sqrt{\dfrac{847}{27}}}+\sqrt[3]{6-\sqrt{\dfrac{847}{27}}} \bigg ]$
$\Leftrightarrow A^3 =12+3 \sqrt[3]{\dfrac{125}{27}} . A=12 +5A $
$\Leftrightarrow A^3-5A-12=0$
$\Leftrightarrow (A-3)(A^2 +3A+4)=0$
Từ đó $A=3$