Binomial Series question help!

Give the binomial expansion, for small x, of (1+x)^(1/3), up to and including the term in x^2
By putting x=1/27 in your expansion, show that 28^(1/3) is approximately equal to 6641/2187

Derive the binomial expansion (i.e. Taylor Series):

f(x) = (1 + x)^(1/3)
f'(x) = 1/3(1 + x)^(-2/3)
f''(x) = -2/9 * (1 + x)*^(-5/3)
-->

f(0) = (1 + 0)^(1/3) = 1
f'(0) = 1/3(1 + 0)^(-2/3) = 1/3
f''(x) = -2/9 * (1 + 0)*^(-5/3) = -2/9
-->

f(x) ~ 1 + (1/3)x - (2/9) / 2! * x²
-->

f(x) ~ 1 + (1/3)x - (1/9) * x²
-->

So by putting in x = 1/27:

f(1/27) ~ 1 + (1/3) * (1/27) - (1/9) * (1/27)²
-->

f(1/27) ~ 1 + 1/(3*27) - 1/(3 * 3 * 27 * 27)
-->

f(1/27) ~ 1 + (3*27)/(3*27 * 3*27) - 1/(3 * 3 * 27 * 27)
-->

f(1/27) ~ 1 + (81 - 1) / 6561 = 1 + 80/6561
-->

f(1/27) ~ (6561 + 80)/6561 = 6641/6561


Now let's look at the original equation:

(1 + 1/27)^(1/3) = ((27 + 1)/27)^(1/3) = (28/27)^(1/3)
--> but this is equal to

28^(1/3) / 27^(1/3) = 28^(1/3) / 3

(since 3*3*3 = 27 --> 27^(1/3) = 3)

28^(1/3) = 3 * (28/27)^(1/3) ~ 3 * 6641/6561
-->

but remember: 6561 = 3*3*27*27
--> if we cancel that with 3 we get:

6561/3 = 2187 OR 3*27*27 = 2187
-->

28^(1/3) = 3 * (28/27)^(1/3) ~ 3 * 6641/6561 = 6641/2187