Given, curve is y=x3−3x ......(i) On differentiating w.r.t. x, we get ‌
dy
dx
=3x2−3 since, tangent is parallel to x -axis. ∴‌‌‌
dy
dx
=0 3x2−3=0⇒x2=1 ⇒‌‌x=±1 ∴ From Eq. (i), when x=1,‌‌y=13−3(1)=−2 when x=−1,‌‌y=(−1)3−3(−1)=2 ∴ Required points are (1,−2) and (−1,2)