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?
 A: \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 $x = -1, 0, 1$
A: To find the intersection between the graph, just find the solutions of
$x^3=x^{1/3}$.
$$x^3=x^{1/3} \Rightarrow x^9-x=x(x-1)(x+1)(x^2+1)(x^4+1)=0 \Rightarrow x=-1,0,1$$
Since $x^3\geq x^{1/3}$ for $x\in[-1,0]$, and $x^{1/3}\geq x^3$ for $x\in[0,1]$, the area between the two graphs is given by
$$\int_{-1}^0 (x^3-x^{1/3})dx+\int_0^1 (x^{1/3}-x^3)dx$$
A: Solve the equation $x^3=\sqrt[3]{x}$:
$$
x^3=x^{\frac13}\implies\\
\left(x^3\right)^3=\left(x^{\frac13}\right)^3\implies\\
x^9=x.
$$
Divide both sides by $x$ and observe that $x=0$ is a solution:
$$
\frac{x^9}{x}=1\implies\\
x^{9-1}=1\implies\\
x^8=1\implies\\
x=\pm1.
$$
So, the solution set consists of three elements: $\{-1,0,1\}$. And those are the $x$ values for your intersection points.
