Find the inverse function of $f(x) = \frac{x^3+3x}{2}$ 
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$

My thought on this are as follows:
I've solved a similar problem where I had to find the inverse of:
$$
g(x) = \sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}
$$
So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and $b = \sqrt[3]{x-\sqrt{x^2-1}}$, then
$y = a + b$ and hence:
$$
y^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = \\
= 2x+3ab(a+b)=2x+3(a+b)
$$
But $y = a + b$ and then
$$
y^3 = 2x + 3y \iff x = \frac{y^3-3y}{2}
$$
This function is very similar to the one in the title. 
The answer says the inverse $f^{-1}(x) = \sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}}, \; x\in \mathbb R$. So knowing the answer it's easy to show that it indeed appears to be the inverse of $f(x)$ using the approach above. But how could I achieve the same result given the inverse function is unknown?
 A: Hint:
Let $$x:=\sqrt[3]u-\frac1{\sqrt[3]u}$$
Then
$$x^3+3x=u-3\sqrt[3]u+\frac3{\sqrt[3]u}-\frac1u+3\sqrt[3]u-\frac3{\sqrt[3]u}=u-\frac1u.$$
Now solve the equation 
$$u-\frac1u=2y$$
for $u$ (it can be reduced to a quadratic one).
In the end,
$$x=\sqrt[3]{u(2y)}-\frac1{\sqrt[3]{u(2y)}}.$$
Notice that the product of the two $u$ roots is $-1$, so that the solution can also be expressed as the sum of the cubic roots.
A: You want to solve $x^3 + 3x - 2y = 0$ as $x = f(y)$. The canonical way to do this is to write $x = a + b$, obtaining
$$a^3 + 3a^2b + 3ab^2 + b^3 + 3(a + b) - 2y = 0$$
The trick is specify that $3ab = -3$. The above equation then reduces to
$$a^3 - b^3 - 2y = 0$$
Which is the same as 
$$a^3 - 1/a^3 - 2y = 0$$
Multiplying by $a^3$ gives
$$a^6 - 2ya^3 - 1 = 0$$
You can now use the quadratic formula to find an expression for $a^3$ in terms of $y$, and then you get three possible complex cube roots for $a$. You then can find $b$ through $3ab = -3$ and then you have three possible $x = a + b$. These are the solutions to your original equation. The simplest of these three will be given by the formula you have there.
