Explain how you can approximate a cube root when the cube is not a perfect cube.

Step-by-step

You can  use Newton's method as follows:

It is an iterative method, meaning that from an approximation (i.e. an approximate solution), we can refine the answer to get a closer approximation.  By repeating the process (iteration), we can get an answer as accurate as we wish.

The basis of the formula is based on

x1=x0-f(x0)/f'(x0)

where

x0 is a given approximation

x1 is a better approximation

f(x) is a given function (for which we need the roots)

f'(x) is the derivative of the function.

Skipping all details and applying directly to find the cube root, we have

N=the number for which the cube root is desired

f(x)=x^3-N

f'(x)=3x^2

and x0 is an initial approximation that we need to provide (from the integer cubes, for example).

Say we need to find the cube-root of 124.

We know that 5^3 = 125, a rather close approximation, but the cube root is slightly less than 5.

To find a better approximation, we apply Newton's method, and calculate mentally:

x0=5, N=124

x1 = x0-(x0^3-124)/(3(x0^2))    [next substitute values]

= 5 - (125-124)/(3(5^2))     [ next simplify ]

= 5-1/75    [next, rearrange to calculate mentally ]

=5-(1/25)/3  [ next, substitute 1/25 = 0.04, 1/3 division can be done mentally ]

= 5 - 0.04/3    [ divide 0.04/3 mentally ]

= 5 - 0.013333    [ subtract ]

= 4.98667

(exact value = 4.98663, first approximation is already quite close)

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50