Philippine Mathematical Olympiad

Solution:

Let $a=\sqrt[3]{5 \sqrt{2}+7}$ and let $b=\sqrt[3]{5 \sqrt{2}-7}$. Note that
$$a^{3}-b^{3}=14 \text{ and } a b=\sqrt[3]{50-49}=1$$
Now, $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)=(a-b)\left[(a-b)^{2}+3 a b\right]$. Letting $x=a-b$, we have the resulting equation $x\left(x^{2}+3\right)=14$.
$$x\left(x^{2}+3\right)=14 \rightarrow x^{3}+3 x-14=0 \rightarrow(x-2)\left(x^{2}+2 x+7\right)=0$$
The roots of $x^{2}+2 x+7$ are not real since its discriminant is $2^{2}-4(7)=-24<0$. Since $x=a-b$ is a real number, then $x=2$.

(Alternatively, the expression is equivalent to $(\sqrt{2}+1)-(\sqrt{2}-1)=2$.)