Area between $x^3$ and $\sqrt[3]{x}$ I'm having some difficult to find the intersection between $x^3$ and $\sqrt[3]{x}$ for calculate the area between them. Could someone help me?

\begin{equation}
x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0
\end{equation}
And so we have $x = 0$ or $x^8 - 1 = 0$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus, 
\begin{equation}
x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0
\end{equation}
Assuming you are seeking Real Solutions only we see that $x^4 + 1 = 0$ has no solutions. For $x^4 - 1$ employ the same identity:
\begin{equation}
(x^4 + 1)(x^4 - 1) = 0 \rightarrow (x^4 + 1)(x^2 + 1)(x^2 - 1) = 0
\end{equation}
As with $x^4 + 1 = 0$ having no Real Solutions we also observe that $x^2 + 1 = 0$ also have no Real Solutions. Thus the only remaining Real Solutions as those that satisfy:
\begin{equation}
x^2 - 1 = 0 \rightarrow x = \pm 1
\end{equation}
As such, the three intersection points of $x^3$ and $x^{\frac{1}{3}}$ occur at $x = -1, 0, 1$
