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The problem asks to expand \(\frac{x+3}{(3 - x)^2}\) using the binomial theorem, as far as the term in \(x^3\).

### Step 1: Rewrite the expression
We need to expand:
\[
\frac{x+3}{(3 - x)^2} = (x+3) \cdot (3 - x)^{-2}
\]
We can now focus on expanding \((3 - x)^{-2}\) using the binomial theorem.

### Step 2: Binomial expansion of \((3 - x)^{-2}\)
The binomial expansion for \((1 - u)^n\) is given by:
\[
(1 - u)^n = \sum_{k=0}^{\infty} \binom{n}{k} (-u)^k
\]
Here, \(u = \frac{x}{3}\) and \(n = -2\). We first rewrite \((3 - x)^{-2}\) as:
\[
(3 - x)^{-2} = 3^{-2}(1 - \frac{x}{3})^{-2}
\]

Now, expand \((1 - \frac{x}{3})^{-2}\) up to the term involving \(x^3\):
\[
(1 - \frac{x}{3})^{-2} = 1 + 2 \cdot \frac{x}{3} + 3 \cdot \left(\frac{x}{3}\right)^2 + 4 \cdot \left(\frac{x}{3}\right)^3 + \dots
\]
Simplifying the terms:
\[
= 1 + \frac{2x}{3} + \frac{3x^2}{9} + \frac{4x^3}{27} + \dots
\]
\[
= 1 + \frac{2x}{3} + \frac{x^2}{3} + \frac{4x^3}{27} + \dots
\]

Now, multiply by \(3^{-2} = \frac{1}{9}\):
\[
(3 - x)^{-2} = \frac{1}{9} \left( 1 + \frac{2x}{3} + \frac{x^2}{3} + \frac{4x^3}{27} + \dots \right)
\]
\[
= \frac{1}{9} + \frac{2x}{27} + \frac{x^2}{27} + \frac{4x^3}{243} + \dots
\]

### Step 3: Multiply by \(x + 3\)
Now, we multiply the expansion by \(x + 3\):
\[
\frac{x+3}{(3 - x)^2} = (x + 3) \cdot \left( \frac{1}{9} + \frac{2x}{27} + \frac{x^2}{27} + \frac{4x^3}{243} \right)
\]

Distribute \(x + 3\) over each term:
\[
= \frac{x + 3}{9} + \frac{(x + 3)2x}{27} + \frac{(x + 3)x^2}{27} + \frac{(x + 3)4x^3}{243}
\]

Now, simplify each term:
1. \(\frac{x + 3}{9} = \frac{x}{9} + \frac{3}{9} = \frac{x}{9} + \frac{1}{3}\)
2. \(\frac{(x + 3)2x}{27} = \frac{2x^2 + 6x}{27}\)
3. \(\frac{(x + 3)x^2}{27} = \frac{x^3 + 3x^2}{27}\)
4. \(\frac{(x + 3)4x^3}{243} = \frac{4x^4 + 12x^3}{243}\)

Now combine terms:
\[
= \frac{x}{9} + \frac{1}{3} + \frac{2x^2 + 6x}{27} + \frac{x^3 + 3x^2}{27} + \frac{4x^4 + 12x^3}{243}
\]

For terms up to \(x^3\), the expansion is:
\[
= \frac{x}{9} + \frac{1}{3} + \frac{2x^2}{27} + \frac{6x}{27} + \frac{x^3}{27} + \frac{3x^2}{27} + \frac{12x^3}{243}
\]

Simplifying further:
\[
= \frac{1}{3} + \frac{x}{9} + \frac{8x^2}{27} + \frac{19x^3}{243}
\]

### Final Result:
\[
\frac{x+3}{(3 - x)^2} \text{ up to the term in } x^3 \text{ is: }
\]
\[
\frac{1}{3} + \frac{x}{9} + \frac{8x^2}{27} + \frac{19x^3}{243}
\]

Would you like further details on any specific part? 

### Related Questions:
1. What is the binomial theorem and how is it applied to negative exponents?
2. Can you expand the series to higher-order terms beyond \(x^3\)?
3. How does the coefficient of a term change based on the binomial expansion?
4. How do you handle fractional exponents in binomial expansions?
5. What are some practical applications of binomial expansions?

**Tip:** Always check your coefficients carefully when applying the binomial expansion, especially when fractional exponents or negative powers are involved.

Using the binomial theorem, expand (x + 3)/(3 - x)^2 as far as the term in x^3.

Learn how to expand the expression (x + 3) / (3 - x)^2 using the binomial theorem. This step-by-step guide covers the binomial expansion up to the term involving x^3, suitable for Grades 11-12.

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