## Calculus: Early Transcendentals (2nd Edition)

The equation is symmetric about the $x$-axis, the $y$-axis and the origin.
$x^{2/3}+y^{2/3}=1$ Check for symmetry about the $y$-axis by substituting $x$ by $-x$ in the given expression and simplifying: $(-x)^{2/3}+y^{2/3}=1$ $\sqrt[3]{(-x)^{2}}+y^{2/3}=1$ $\sqrt[3]{x^{2}}+y^{2/3}=1$ $x^{2/3}+y^{2/3}=1$ Since substituting $x$ by $-x$ yielded an equivalent expression, the equation is symmetric about the $y$-axis Check for symmetry about the $x$-axis by substituting $y$ by $-y$ in the given expression and simplifying: $x^{2/3}+(-y)^{2/3}=1$ $x^{2/3}+\sqrt[3]{(-y)^{2}}=1$ $x^{2/3}+\sqrt[3]{y^{2}}=1$ $x^{2/3}+y^{2/3}=1$ Since substituting $y$ by $-y$ yielded an equivalent expression, the equation is symmetric about the $x$-axis Since the equation is symmetric to both the $x$ and $y$-axis, it is also symmetric about the origin. The graph is shown in the answer section.