Let's simplify the given expression step by step:

$\frac{{(12 - 4\sqrt{2} \cdot 3) \cdot (4\sqrt{2} \cdot 3 - 4)}}{2}$

Step 1: Simplify the terms inside the square roots.

First, evaluate the multiplication of $\sqrt{2}$ and 3 in both parts:

• $4\sqrt{2} \cdot 3 = 4 \cdot 3 \cdot \sqrt{2} = 12\sqrt{2}$
• $4\sqrt{2} \cdot 3 = 12\sqrt{2}$ (same term again)

Thus, the expression becomes:

$\frac{{(12 - 4 \cdot 12\sqrt{2}) \cdot (12\sqrt{2} - 4)}}{2}$

Step 2: Distribute the terms.

• $12 - 4 \cdot 12\sqrt{2} = 12 - 48\sqrt{2}$
• $12\sqrt{2} - 4$ remains as it is.

Thus, the expression is now:

$\frac{{(12 - 48\sqrt{2}) \cdot (12\sqrt{2} - 4)}}{2}$

Step 3: Multiply the binomials.

Now, distribute the terms in $(12 - 48\sqrt{2})$ and $(12\sqrt{2} - 4)$:

$= 12 \cdot 12\sqrt{2} + 12 \cdot (-4) - 48\sqrt{2} \cdot 12\sqrt{2} - 48\sqrt{2} \cdot (-4)$

Let’s simplify each term:

• $12 \cdot 12\sqrt{2} = 144\sqrt{2}$
• $12 \cdot (-4) = -48$
• $-48\sqrt{2} \cdot 12\sqrt{2} = -48 \cdot 12 \cdot 2 = -1152$
• $-48\sqrt{2} \cdot (-4) = 192\sqrt{2}$

Now, the expression becomes:

$144\sqrt{2} - 48 - 1152 + 192\sqrt{2}$

Step 4: Combine like terms.

Now, group the $\sqrt{2}$-terms and constant terms:

• $144\sqrt{2} + 192\sqrt{2} = 336\sqrt{2}$
• $-48 - 1152 = -1200$

So the simplified expression becomes:

$\frac{{336\sqrt{2} - 1200}}{2}$

Step 5: Simplify the final expression.

Now, divide each term by 2:

$= 168\sqrt{2} - 600$

Final Answer:

$168\sqrt{2} - 600$

Would you like any additional details or clarification? Here are 5 related questions:

1. How would you solve similar expressions involving square roots and multiplication?
2. How do you handle negative signs when distributing terms in binomials?
3. What is the significance of simplifying expressions involving radicals in algebra?
4. How would you approach an expression with cube roots instead of square roots?
5. What are the properties of square roots that simplify such expressions?

Tip: Always check if you can simplify square root expressions before performing operations—this can often make the problem easier to handle!