2016/062) If $x= 2 + \sqrt[3]{2^2} + \sqrt[3]{2}$ then the value of $x^3 - 6x^2+6x$ is

we have  $x-  2 =  \sqrt[3]{2^2} + \sqrt[3]{2}$
hence $(x-2)^3 = 4 + 2 + 3 * 2 *(\sqrt[3]{2^2} + \sqrt[3]{2}) = 6 + 3 * 2 (x-2) = 6x - 6$
Hence $x^3 - 6x^2 + 12x -8 = 6x-6 $ or $x^3-6x^2+6x = 2$