Q2014/100) show that $\sqrt[3]{45 + 29\sqrt2} + \sqrt[3]{45 - 29\sqrt2}$ is rational

it is as below
 $x=\sqrt[3]{45 + 29\sqrt2} +  \sqrt[3]{45 - 29\sqrt2}$
or
$x-\sqrt[3]{45 + 29\sqrt2} -  \sqrt[3]{45 - 29\sqrt2}= 0$

using $a+b+c = 0 => a^3+b^3+ c^3 = 3abc$
we get
$x^3-(45 + 29\sqrt2) - (45 - 29\sqrt2)= 3 x\sqrt[3]{(45 + 29\sqrt2)(45 - 29\sqrt2)}$
or
$x^3-90= 3x\sqrt[3]{45^2 - 2 * 29^2}$
or $x^3-90 = 21x$
or $x^3 - 21x - 90 = 0$
or $(x-6)(x^2+6x+ 15) = 0$
has one real root = 6 and 2 complex roots

hence given expression = 6 which is real