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

Question

I've seen another equation that I have to solve for $x$. $$\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=\sqrt[3]{5}$$ Hint me how I must simplify it and then solve it. I don't have any ideas! :(

Answer 1

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

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

Cube both sides

Answer 2

HINT

Set $a=\sqrt[3]{1+\sqrt{x}}$ and $b=\sqrt[3]{1-\sqrt{x}}$ and note that

$$\begin{align} a+b&=\sqrt[3]{5}\\ a^3+b^3&=2 \end{align}$$

Further use the binomial expansion of $(a+b)^3$ to get

$$(a+b)^3=a^3+b^3+3ab(a+b)$$

The term $ab$ can be computed by using the first relation. Can you take it from here?