Cubes

Let $\alpha = \sqrt [3] {1+\sqrt{x}}$ and $\beta = \sqrt [3] {1-\sqrt{x}}$ . Then:

$(\alpha + \beta)^3 = 5$

$\alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 = 5$

Note that: $\quad \alpha^3 + \beta^3 = 1 + \sqrt{x} + 1 - \sqrt{x} = 2$

Therefore, $\quad 2 + 3\alpha\beta ( \alpha + \beta) = 5 \quad \Rightarrow 3\alpha\beta ( \alpha + \beta) = 3 \quad \Rightarrow \alpha\beta ( \alpha + \beta) = 1$

Cubing throughout: $\quad \alpha^3\beta^3(\alpha+\beta)^3 = 1\quad \Rightarrow (1 + \sqrt{x})(1-\sqrt{x})(5) = 1$

$\Rightarrow 5(1-x) = 1 \quad \Rightarrow 1 - x = \dfrac {1}{5}\quad \Rightarrow x = \frac {4}{5} = \frac {a}{b} \quad \Rightarrow a + b = \boxed {9}$

$\sqrt[3]{\frac{1 + \sqrt{x}}{5}} + \sqrt[3]{\frac{1 - \sqrt{x}}{5}} = 1$

Cubing

$\frac{1 + \sqrt{x}}{5} + \frac{1 - \sqrt{x}}{5} + 3\times\sqrt[3]{\frac{1 - x}{25}} = 1$

$3\times\sqrt[3]{\frac{1 - x}{25}} = \frac{3}{5}$

$\frac{1 - x}{25} = \frac{1}{125}$

$x = \frac{4}{5}$

we have

$\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}+\sqrt[3]{-5}=0$

This should instantly motivate us to use the famous equality:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

and if, $a+b+c=0$ ; we have $a^{3}+b^{3}+c^{3}=3abc$

plugging in $a=\sqrt[3]{1+\sqrt{x}}$ ; .. $b=\sqrt[3]{1-\sqrt{x}}$ and $c=\sqrt[3]{-5}$

we have:

$1+\sqrt{x}+1-\sqrt{x}-5=3\sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})(-5)}$

Simplifying,

$\large{x=\frac{4}{5}}$

So, our answer: $\LARGE{\boxed{9}}$

@megh choksi .. what do you think? :D

I think the simler method is the standard one in which you cube the equation, may there be some concept which guaranties a more clever approach?