Complex equality

$\begin{aligned} \sqrt[3]{\sqrt[3]{5\sqrt2+x}+\sqrt[3]{5\sqrt2-x}} & = \sqrt 2 & \small \color{#3D99F6} \text{Cubing both sides} \\ \sqrt[3]{5\sqrt2+x}+\sqrt[3]{5\sqrt2-x} & = 2 \sqrt 2 & \small \color{#3D99F6} \text{Cubing again} \\ 5\sqrt2 + x + 3\sqrt[3]{\left(5\sqrt2+x\right) \left(5\sqrt2-x\right)}\left({\color{#3D99F6}\sqrt[3]{5\sqrt2+x}+\sqrt[3]{5\sqrt2-x}}\right) +5\sqrt2-x & = 16 \sqrt 2 \\ 10\sqrt2 + 3\sqrt[3]{50-x^2}\left({\color{#3D99F6}2\sqrt 2}\right) & = 16 \sqrt 2 \\ \sqrt[3]{50-x^2} & = 1 & \small \color{#3D99F6} \text{Cubing again} \\ 50-x^2 & = 1 \\ x^2 & = 49 \\ \implies x & = \pm 7 \end{aligned}$

And the answer is $x = \boxed{7}$ .

Let $U = \sqrt[3] {5\sqrt{2} +x}$ and $V = \sqrt[3] {5\sqrt{2} -x}$ .

We observe that

$\begin{aligned} & U^3 + V^3 = 10 \sqrt{2} \\ & UV = \sqrt[3] {50 - x^2} \end{aligned}$

Using the formula $x^3 + y^3 = (x+y)^3 - 3xy (x+y)$ , we get

$\begin{aligned} & U^3 + V^3 = 10 \sqrt{2} \\ \implies & (U+V)^3 - 3UV (U+V) = 10 \sqrt{2} \\ \implies & (U+V)^3 - 3 \sqrt[3] {50-x^2} (U+V) = 10 \sqrt{2} \end{aligned}$

From the given equality

$\sqrt[3]{ \sqrt[3]{5 \sqrt{2} + x} + \sqrt[3]{5 \sqrt{2} - x} } = \sqrt{2}$

We have that

$\sqrt[3] {U+V} = \sqrt{2} \implies U+V = 2^{{3}/{2}}$

Thus

$\begin{aligned} & (U+V)^3 - 3 \sqrt[3] {50-x^2} (U+V) = 10 \sqrt{2} \\ \implies & 2^{{9}/{2}} - 3 \sqrt[3] {50-x^2} \cdot 2^{{3}/{2}} = 10 \cdot 2^{{1}/{2}} \\ \implies & 2^4 - 3 \sqrt[3] {50-x^2} \cdot 2 = 10 \\ \implies & 6 \sqrt[3] {50-x^2} = 6 \\ \implies & 50-x^2 = 1^3 \\ \implies & x^2 = 49 \\ \implies & x = \pm 7 \end{aligned}$

So the positive value of $x$ would be $\boxed{x=7}$ .

Manipulating the equation we have

$\begin{aligned} \sqrt[3]{\sqrt[3]{5\sqrt{2}+x} + \sqrt[3]{5\sqrt{2}-x}} &= \sqrt{2} \\ \sqrt[3]{5\sqrt{2}+x} + \sqrt[3]{5\sqrt{2}-x} &= 2\sqrt{2} \\ \sqrt[3]{5\sqrt{2}+x} + \sqrt[3]{5\sqrt{2}-x} - 2\sqrt{2} &= 0 \end{aligned}$

Using the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ , when $a+b+c=0$ we have $a^3+b^3+c^3-3abc=0$ .

Let $a=\sqrt[3]{5\sqrt{2}+x}, b=\sqrt[3]{5\sqrt{2}-x} , c=-2\sqrt{2}$ , we have

$\begin{aligned} a^3+b^3+c^3-3abc &= 0 \\ (5\sqrt{2}+x)+(5\sqrt{2}-x) -16\sqrt{2} + 6\sqrt{2}\big( \sqrt[3]{(5\sqrt{2}+x)(5\sqrt{2}-x)}\big )&= 0 \\ 6\sqrt{2} \big ( \sqrt[3]{50-x^2} - 1\big) &= 0 \\ \sqrt[3]{50-x^2} - 1 &= 0 \\ \sqrt[3]{50-x^2} &= 1 \\ 50-x^2 &= 1 \\ x^2 &= 49 \\ x &= \pm 7 \end{aligned}$