Question

How do you solve `root(3) { 5.67}`?

Answer 1

`root(3)(5.67) ~~ 1.783176586`

Explanation:

You can calculate approximations to the cube root of a number using Newton's method:

Given a number `n` whose cube root you wish to find and an approximation `a_i` to `root(3)(n)`, a better approximation is:

`a_(i+1) = a_i - (a_i^3-n)/(3a_i^2) = (2a_i^3+n)/(3a_i^2)`

Repeat to get successively better approximations.

For our example, with `n=5.67`, let `a_0 = 2`

Then:

`a_0 = 2`

`a_1 = (2a_0^3+n)/(3a_0^2) = (2(color(blue)(2))^3+5.67)/(3(color(blue)(2))^2) = (16+5.67)/12 = 1.8058bar(3)`

`a_2 = (2a_1^3+n)/(3a_1^2) = (2(color(blue)(1.8058bar(3)))^3+5.67)/(3(color(blue)(1.8058bar(3)))^2) ~~ 1.783459658`

`a_3 = (2a_2^3+n)/(3a_2^2) = (2(color(blue)(1.783459658))^3+5.67)/(3(color(blue)(1.783459658))^2) ~~ 1.783176631`

`a_4 = (2a_3^3+n)/(3a_3^2) = (2(color(blue)(1.783176631))^3+5.67)/(3(color(blue)(1.783176631))^2) ~~ 1.783176586`