What is the equation of the normal line to the curve y=cube root(x^2-1) at x=3?

We know that the product of slopes of two perpendicular
lines is -1. Therefore, we'll determine the slope of the tangent line to the given
curve, at x = 3.

We'll find the slope m using
derivative:

dy/dx = (1/3)*[(x^2 -
1)^(-2/3)]*(2x)

We'll calculate the slope at
x=3

dy/dx = m = (1/3)*[(9 -
1)^(-2/3)]*(6)

m = 2/8^(2/3)

m
= 2/4

m = 1/2

The slope of the
normal line is m1:

m*m1 = -1=>  m1 = -1/(1/2) =
-2

Now, we'll calculate the y coordinate at
x=3:

y = cube root(9-1) = cube root 8 =
2

The equation of a line that passes through a point and it
has a slope is:

y - y1 = m1*(x -
x1)

Comparing, we'll get:

y -
2 = -2(x-3)

Therefore, the equation of normal
line is:  y = -2x + 8.