We are given the expression \((X+2)(X-3)(X-3) + 64\), and our goal is to factorize it. Let's go step by step:

### Step 1: Simplify the product \((X - 3)(X - 3)\)
This is a perfect square:

\[
(X - 3)(X - 3) = (X - 3)^2 = X^2 - 6X + 9
\]

### Step 2: Expand \((X + 2)(X^2 - 6X + 9)\)
Now we expand the expression \((X + 2)(X^2 - 6X + 9)\) by distributing \(X + 2\) over the trinomial:

\[
(X + 2)(X^2 - 6X + 9) = X(X^2 - 6X + 9) + 2(X^2 - 6X + 9)
\]

First, expand each part:
- \(X(X^2 - 6X + 9) = X^3 - 6X^2 + 9X\)
- \(2(X^2 - 6X + 9) = 2X^2 - 12X + 18\)

Now, combine these terms:
\[
X^3 - 6X^2 + 9X + 2X^2 - 12X + 18 = X^3 - 4X^2 - 3X + 18
\]

### Step 3: Add 64
Now, we add 64 to the expanded expression:
\[
X^3 - 4X^2 - 3X + 18 + 64 = X^3 - 4X^2 - 3X + 82
\]

### Step 4: Factor the resulting expression
The expression is now \(X^3 - 4X^2 - 3X + 82\). We attempt factoring by grouping or other methods, but this particular cubic expression doesn't factor easily over the integers. However, it is the simplified form of the original expression.

Thus, the fully simplified form is:
\[
X^3 - 4X^2 - 3X + 82
\]

This is the factored and simplified result.

Would you like further details or clarification?

Here are 5 related questions to further your understanding:
1. How do you factor cubic polynomials in general?
2. Can you explain what a perfect square trinomial is?
3. What techniques can be used to factor higher-degree polynomials?
4. What are the possible methods to solve cubic equations?
5. What role does the distributive property play in expanding algebraic expressions?

**Tip**: When factorizing expressions, always look for common patterns like perfect squares or cubes to simplify the process.

Learn how to factorize the algebraic expression (X+2)(X-3)(X-3) + 64 using polynomial factorization techniques. This solution walks through expanding perfect square trinomials and simplifying cubic expressions. Suitable for high school students in Grades 9-12.

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