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The image contains the following mathematical problem:

**Part (i):**
Find the binomial expansion of \( (27 + 6x)^{\frac{1}{3}} \) up to and including the term in \( x^3 \), simplifying the coefficients.

**Part (ii):**
Given that \( \sqrt[3]{3} = \frac{3}{\sqrt[3]{9}} \), use your binomial expansion in part (i) to obtain the value of \( \sqrt[3]{3} \), giving the answer to five significant places.

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### Part (i): Binomial Expansion

The expression given is \( (27 + 6x)^{\frac{1}{3}} \).

To simplify the binomial expansion, we can factor out the 27:

\[
(27 + 6x)^{\frac{1}{3}} = 27^{\frac{1}{3}} \left( 1 + \frac{6x}{27} \right)^{\frac{1}{3}} = 3 \left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}}
\]

Now, we use the binomial expansion formula for \( (1 + u)^n \) where \( u = \frac{2x}{9} \) and \( n = \frac{1}{3} \):

\[
(1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \dots
\]

Expanding \( \left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}} \):

\[
\left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}} = 1 + \frac{1}{3} \cdot \frac{2x}{9} + \frac{\frac{1}{3} \cdot \left( \frac{1}{3} - 1 \right)}{2!} \cdot \left( \frac{2x}{9} \right)^2 + \dots
\]

This simplifies to:

\[
= 1 + \frac{2x}{27} - \frac{2x^2}{243} + \dots
\]

Thus, multiplying by 3 (from the earlier factoring out of 27):

\[
3 \left( 1 + \frac{2x}{27} - \frac{2x^2}{243} + \dots \right) = 3 + \frac{2x}{9} - \frac{2x^2}{81} + \dots
\]

Including up to the \( x^3 \) term, we have:

\[
3 + \frac{2x}{9} - \frac{2x^2}{81} + \dots
\]

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### Part (ii): Using Binomial Expansion to Estimate \( \sqrt[3]{3} \)

We are given \( \sqrt[3]{3} = \frac{3}{\sqrt[3]{9}} \).

We know that \( \sqrt[3]{9} = (3^2)^{1/3} = 3^{2/3} \).

To estimate \( \sqrt[3]{3} \), we can use the binomial expansion we found in part (i). By setting \( x = 1 \), we substitute this into the expansion from part (i):

\[
\sqrt[3]{9} \approx 3 + \frac{2 \cdot 1}{9} - \frac{2 \cdot 1^2}{81} = 3 + \frac{2}{9} - \frac{2}{81} = 3 + 0.2222 - 0.0247 = 3.1975
\]

Therefore,

\[
\sqrt[3]{3} = \frac{3}{\sqrt[3]{9}} \approx \frac{3}{3.1975} = 0.9383
\]

Finally, rounding to five significant places:

\[
\sqrt[3]{3} \approx 0.93828
\]

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Would you like more details on any step? 

Here are some related questions:

1. How do we derive the general binomial expansion for fractional powers?
2. Why is binomial expansion useful for approximations like these?
3. What happens if we include more terms in the binomial expansion?
4. How can we refine the approximation further using calculus methods?
5. Can we use this method to approximate other roots, such as \( \sqrt[4]{3} \)?

**Tip:** When using binomial expansions, remember to watch for how many terms you need to include for the desired accuracy.

Find the binomial expansion of (27 + 6x)^(1/3) up to and including the term in x^3, simplifying the coefficients. Then, given that cube root of 3 = 3/cube root of 9, use the binomial expansion to estimate the cube root of 3, giving the answer to five significant places.

Learn how to expand (27 + 6x)^(1/3) using binomial expansion and apply it to estimate the cube root of 3. This detailed solution simplifies the coefficients and demonstrates how to approximate values using the expansion, rounding results to five significant places.

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