The problem requires us to find the first four terms of the binomial expansion of $\sqrt[3]{1+x}$ and then use this expansion to evaluate $\sqrt[3]{9}$ to three decimal places.

1. Problem Description

The problem requires us to find the first four terms of the binomial expansion of $\sqrt[3]{1+x}$ and then use this expansion to evaluate $\sqrt[3]{9}$ to three decimal places.

2. Solution Steps

First, we find the first four terms of the binomial expansion of $(1+x)^{1/3}$ using the binomial theorem. The general formula for the binomial expansion is:

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$

In our case, $n = 1/3$. So the first four terms are:

Term 1: $1$

Term 2: $\frac{1}{3}x$

Term 3: $\frac{(\frac{1}{3})(\frac{1}{3}-1)}{2!}x^2 = \frac{(\frac{1}{3})(-\frac{2}{3})}{2}x^2 = -\frac{1}{9}x^2$

Term 4: $\frac{(\frac{1}{3})(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}x^3 = \frac{(\frac{1}{3})(-\frac{2}{3})(-\frac{5}{3})}{6}x^3 = \frac{10}{162}x^3 = \frac{5}{81}x^3$

Therefore, the first four terms of the expansion are:

$(1+x)^{1/3} \approx 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3$

Now, we want to evaluate $\sqrt[3]{9}$. We can rewrite this as:

$\sqrt[3]{9} = \sqrt[3]{8+1} = \sqrt[3]{8(1+\frac{1}{8})} = 2\sqrt[3]{1+\frac{1}{8}}$

Now, we can use the binomial expansion we found with $x = \frac{1}{8}$:

$2\sqrt[3]{1+\frac{1}{8}} \approx 2\left(1 + \frac{1}{3}\left(\frac{1}{8}\right) - \frac{1}{9}\left(\frac{1}{8}\right)^2 + \frac{5}{81}\left(\frac{1}{8}\right)^3\right)$

$= 2\left(1 + \frac{1}{24} - \frac{1}{9(64)} + \frac{5}{81(512)}\right)$

$= 2\left(1 + \frac{1}{24} - \frac{1}{576} + \frac{5}{41472}\right)$

$= 2\left(1 + 0.041666... - 0.001736... + 0.000120...\right)$

$= 2\left(1.041666 - 0.001736 + 0.000120\right)$

$= 2(1.04005)$

$= 2.08010$

Rounding to three decimal places, we get $2.080$.

3. Final Answer

2. 080