\begin{align} & \frac{2(a+b)x+2ab}{4x+a+b}={{(\frac{{{a}^{\frac{1}{3}}}+{{b}^{\frac{1}{3}}}}{2})}^{3}} \\ & \Leftrightarrow 8\left( 2(a+b)x+2ab \right)=\left( a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)(4x+a+b) \\ & \Leftrightarrow 16(a+b)x+16ab=\left( 4(a+b)+12\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)x+(a+b)\left( a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right) \\ & \Leftrightarrow 12\left( a+b-\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)x=(a+b)\left( a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)-16ab \\ & \Leftrightarrow x=\frac{(a+b)\left( a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)-16ab}{12\left( a+b-\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b}) \right)} \end{align}

Another solution. 1) $f(x) = m - \frac{n}{x + k}$, where $m = \frac{a + b}{2}, n = \frac{(a + b)^2}{2} - 2ab, k$ all $ > 0$. 2) For positive x-es $f(x)$ assumes range $(f(0); m)$, so we need to show, that $\frac{2ab}{a + b} < (\frac{a^{1/3} + b^{1/3}}{2})^3 < \frac{a + b}{2}$. 3) Left one is double AM-GM, right one is power mean. 4) f(x) is injective since $a \neq b$, so such x is unique.

I think that you forgot about restriction on $a$ and $b$ that $a+b \neq a+b-\sqrt[3]{ab}(\sqrt[3]{a} +\sqrt[3]{b}))$.