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The equation I have is

$$\Large x^{\frac23} + y^{\frac23} = 3^{\frac23} $$

I know what the graph looks like, but I don't know how I would find points other than the intercepts mathematically. How would I do that?

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  • $\begingroup$ I personally don't understand what it is you're asking - please clarify. $\endgroup$ – lemon Jul 17 '14 at 0:49
  • $\begingroup$ If vyou are curious about the $y$'s that correxpond to a particular value of $x$, such as $x=0.4$, plug in and solve for $y$. You might be interested also in a question like when is $y=x$, i.e. where does the curve meet the line $y=x$. One can calculate, just put $y=x$ and solve. $\endgroup$ – André Nicolas Jul 17 '14 at 0:53
  • $\begingroup$ I think Kekker's problem lies in the understanding of fractional exponents. $\endgroup$ – beanshadow Jul 17 '14 at 0:56
  • $\begingroup$ This is an astroid, a special case of a superellipse. $\endgroup$ – Lucian Jul 17 '14 at 1:20
  • $\begingroup$ I understand what fractional exponents are, and I know the shape of the graph. I just don't understand how to undo a fractional exponent in this equation so as to get an answer. For example, I plug in 1 for x, so I can find the y coordinate for x=1 for my graph. $$\Large 1^{\frac23} + y^{\frac23} = 3^{\frac23} $$ That I can simplify to $$\Large 1 + y^{\frac23} = 3^{\frac23} $$ How would I simplify that further? $\endgroup$ – Kekker Jul 17 '14 at 21:59
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Try $\begin{cases} x=3\cos^3\theta\\ y=3\sin^3\theta\\ \end{cases}$ or, for rational solutions, $\begin{cases} x=3\left({{2t}\over{t^2+1}}\right)^3\\ y=3\left({{t^2-1}\over{t^2+1}}\right)^3\\ \end{cases}$

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  • $\begingroup$ Perhaps this would lead to a rational parametrization of the curve, using the Weierstrass substitution. (+1) $\endgroup$ – coffeemath Jul 17 '14 at 5:43
  • $\begingroup$ I like your second formulation because it allows quick and easy hand computation of loads of rational points. $\endgroup$ – Lubin Jul 17 '14 at 13:04

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