Find the positive integer x such that this equation holds good $\sqrt[3]{20+x\sqrt{2}}+\sqrt[3]{20-x\sqrt{2}}=4.$

Let us take the cube of the original relation and use the identity

$(a+b{)}^3=a^3+b^3+3ab(a+b).$
We obtain
$64=40+3\sqrt[3]{400-2x^2}\cdot 4$

and then $x^2=196$, with the solutions $x=\pm 14$ We conclude that the equation has two solutions, namely 14 and -14 .
Let us give an alternative approach: we introduce two new variables  $a=\sqrt[3]{20+x\sqrt{2}}\text{and}b=\sqrt[3]{20-x\sqrt{2}}$  The equation becomes a+b=4, but there is a hidden relationship between a and b. Indeed, taking the cube of a and b we obtain 

$a^3=20+x\sqrt{2},b^3=20-x\sqrt{2},$

hence by adding these two relations  $\mathrm{s}{a}^3+b^3=40.$ On the other hand, since a+b=4, we have 

$a^3+b^3=(a+b)(a^2-ab+b^2)=4\left(\right(a+b{)}^2-3ab)=4(16-3ab)=64-12ab$ thus
$ab=\frac{64-40}{12}=2$
On the other hand $ab=\sqrt[3]{400-2x^2}$ thus $400-2x^2=8\text{and then}{x}^2=196$ and  $x=\pm 14$ It follows that the solutions of the equation are 14 and -14 .