Ta có $\frac{847}{27} = 36-\frac{125}{27}$
Sử dụng hằng đẳng thức
$(a+b)^3 = a^3+b^3+3ab(a+b)$
Đặt $A = \sqrt[3]{6+\sqrt{\frac{847}{27}}}+\sqrt[3]{6-\sqrt{\frac{847}{27}}}$
$A^3 = 12+3\sqrt[3]{\left(6+\sqrt{36-\frac{125}{27}}\right)\left(6-\sqrt{36-\frac{125}{27}}\right)}A$
$A^3 = 12+3\sqrt[3]{\frac{125}{27}}A$
$A^3 = 12+3\frac{5}{3}A$
$A^3 -5A-12 = 0$
$(A-3)(A^2+3A+4)=0$
hay $A =3$
Vậy
$\sqrt[3]{6+\sqrt{\frac{847}{27}}}+\sqrt[3]{6-\sqrt{\frac{847}{27}}} =3 $