# How do you find a one-decimal place approximation for #-root3 35#?

##### 1 Answer

Oct 12, 2015

Consider

#### Explanation:

We know that

#(3+t)^3=3^3+3*3^2t+3*3t^2+t^3#

#=27+27t+9t^2+t^3#

#~~27+27t# for small values of#t#

So approximating...

#35=27+27t#

Hence

#t=(35-27)/27=8/27~~0.3#

So

More generally, to find the cube root of a number

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

This is a form of Newton Raphson method. Given a function

#a_(i+1) = a_i - f(a_i)/(f'(a_i))#

In our case