Đặt `b=`$(\sqrt[3]{2}+1)(\sqrt[3]{\frac{\sqrt[3]{2}-1}{3}})$
Đặt $\sqrt[3]{2}=y⇒y^3=2$
$⇒b=(y+1)(\sqrt[3]{\frac{y-1}{3}})$
$b=\sqrt[3]{(y+1)^3\frac{y-1}{3}}$
$b=\sqrt[3]{[(y^3+1+3y(y+1)]\frac{y-1}{3}}$
$b=\sqrt[3]{[(2+1+3y(y+1)]\frac{y-1}{3}}$
$b=\sqrt[3]{[(3+3y(y+1)]\frac{y-1}{3}}$
$b=\sqrt[3]{[3(1+y(y+1)].\frac{1}{3}.(y-1)}$
$b=\sqrt[3]{(1+y^2+y).(y-1)}$
$b=\sqrt[3]{(y-1)(y^2+y+1)}$
$b=\sqrt[3]{y^3-1}$
$b=\sqrt[3]{2-1}$
$b=\sqrt[3]{1}$
$b=1$
`⇒a=3+b=3+1=4.`
