2022/040) If $x = 3 + \sqrt[3]3 + \sqrt[3]{3^2}$ then show that $x^3 - 9x^2+ 18x -12 = 0$

 From the given condition we have

$(x-3) = \sqrt[3]3 + \sqrt[3]{3^2}$

squaring both sides

$(x-3)^3 = (\sqrt[3]3)^3 + ( \sqrt[3]{3^2})^2 + 3  \sqrt[3]{3} \sqrt[3]{3^2}(  \sqrt[3]{3} +  \sqrt[3]{3^2})$ using $(a+b)^3 = a ^3 + b^3 + 3ab(a+)$

or $(x-3)^3 = 3 + 9 + 9 ( x-3)$ using $(  \sqrt[3]{3^2} +  \sqrt[3]{3^2} = x- 3$

or $(x^3 - 9x^2 + 27 x - 27 = 12 + 9x - 27 = 9x + 12$

or $x^3 - 9x^2+ 18x -12 = 0$