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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?

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  • $\begingroup$ For both $x=0$ and $x=1$, we have $x^3 = x^{1/3}$. $\endgroup$ – mjw Jul 1 at 4:34
  • $\begingroup$ For one of the roots of $x^{1/3}$, we also have $x=-1$, which will be a point of intersection on the graph $x$ vs $f(x)$ with $x$ and $f(x)$ real. $\endgroup$ – mjw Jul 1 at 4:34
  • $\begingroup$ But how do you get this result? $\endgroup$ – Mycroft Jul 1 at 4:35
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    $\begingroup$ How to get what result: $\{-1,0,1\}$? $\endgroup$ – mjw Jul 1 at 4:36
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    $\begingroup$ This is not how I solved it (I just saw that $x^3$ and $x^{1/3}$ are inverses so they are symmetric with respect to the line $x=y$. Anyway, cube both sides: $x^9-x=0$. You can factor this. $x(x^8-1)=0$. This has roots $x=0$ and the "eigth roots of unity" of which two are real: $x\in\{-1,1\}$. $\endgroup$ – mjw Jul 1 at 4:40
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\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 $x = -1, 0, 1$

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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 $$\int_{-1}^0 (x^3-x^{1/3})dx+\int_0^1 (x^{1/3}-x^3)dx$$

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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.

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  • $\begingroup$ Nice. I like it. $\endgroup$ – mjw Jul 1 at 4:46

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