Find the inverse of the cubic function What is the resulting equation when $y=x^3 + 2x^2$ is reflected in the line $y=x$ ? 
I have tried and tried and am unable to come up with the answer. 
The furthest I was able to get without making any mistakes or getting confused was $x= y^3 + 2y^2$. What am I supposed to do after that step? 

The general solution to the cubic equation
$$a x^3 + b x^2 + c x + d = 0$$
can be written
$$x = -\frac{1}{3 a} \left( b + \sigma C - \sigma \frac{\Delta_0}{C} \right)$$
where
$$
\Delta_0 = b^2 - 3 a c \\
\Delta_1 = 2 b^3 - 9 a b c + 27 a^2 d \\
C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}} \\
\sigma = 1, \frac{1}{2} \pm i \frac{\sqrt{3}}{2}
$$
Note that if you don't care about complex solutions, then you can just ignore $\sigma$.
Comparing this to our equation
$$x = y^3 + 2 y^2 \\
y^3 + 2 y^2 - x = 0$$
we see that all we need to to turn the general form at the top into our problem is substitute $y$ for $x$, $1$ for $a$, $2$ for $b$, $0$ for $c$, and $-x$ for $d$. Plugging these values into the formula and simplifying yields our solution:
$$
\Delta_0 = (2)^2 - 3 (1) (0) = 4 \\
\Delta_1 = 2 (2)^3 - 9 (1) (2) (0) + 27 (1)^2 (-x) = -27 x + 16 \\
C = \sqrt[3]{\frac{(-27 x + 16) \pm \sqrt{(-27 x + 16)^2 - 4 (4)^3}}{2}} = \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \\
y = -\frac{1}{3 (1)} \left((2) + \sigma C - \sigma \frac{(4)}{C}\right) \\
y = -\frac{2}{3} - \frac{\sigma}{3} \left(\textstyle \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \displaystyle - \frac{4}{\sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}}}\right)
$$
