91Ó°ÊÓ

We have given the following equation :-

$y=x^3-x$.

We have to check the symmetry of this equation with respect to x-axis, y-axis and the origin.

The given equation is :-

$y=x^3-x$.

We know that a graph is symmetrical about x-axis, if a point $\left(x,y\right)$lies on graph, then $\left(x,-y\right)$is also lies on graph.

So to check symmetry about x-axis, change $y$by$-y$inthe given equation, then we have :-

$-y=x^3-x$.

This equation is not same as the given equation.

So we can conclude that the given equation is not symmetric about x-axis.

The given equation is :-

$y=x^3-x$.

We know that a graph is symmetrical about y-axis, if a point $\left(x,y\right)$lies on graph, then $\left(-x,y\right)$is also lies on graph.

So to check symmetry about y-axis, change $x$by $-x$in the given equation, then we have :-

$$\begin{gathered} y={\left(-x\right)}^3-\left(-x\right) \\ ⇒y=-x^3+x \end{gathered}$$

This equation is not same as the given equation.

So we can conclude that the given equation is not symmetric about y-axis.

The given equation is :-

$y=x^3-x$.

We know that a graph is symmetrical about origin, if a point $\left(x,y\right)$lies on graph, then $\left(-x,-y\right)$is also lies on graph.

So to check symmetry about origin, change $x$by $-x$and $y$by $-y$in the given equation, then we have :-

$$\begin{gathered} -y={\left(-x\right)}^3-\left(-x\right) \\ ⇒-y=-x^3+x \\ ⇒-y=-\left(x^3-x\right) \\ ⇒y=x^3-x \end{gathered}$$

The resulting equation is same as the given equation.

So we can say that the given equation is symmetric about origin.