Find the Product ( square root of 6-3)( square root of 6+3)

$(\sqrt{6} - 3) (\sqrt{6} + 3)$

Step 1

Expand $(\sqrt{6} - 3) (\sqrt{6} + 3)$ using the FOIL Method.

Apply the distributive property.

$\sqrt{6} (\sqrt{6} + 3) - 3 (\sqrt{6} + 3)$

Apply the distributive property.

$\sqrt{6} \sqrt{6} + \sqrt{6} \cdot 3 - 3 (\sqrt{6} + 3)$

Apply the distributive property.

$\sqrt{6} \sqrt{6} + \sqrt{6} \cdot 3 - 3 \sqrt{6} - 3 \cdot 3$

Step 2

Simplify and combine like terms.

Simplify each term.

Combine using the product rule for radicals.

$\sqrt{6 \cdot 6} + \sqrt{6} \cdot 3 - 3 \sqrt{6} - 3 \cdot 3$

Multiply $6$ by $6$ .

$\sqrt{36} + \sqrt{6} \cdot 3 - 3 \sqrt{6} - 3 \cdot 3$

Rewrite $36$ as $6^{2}$ .

$\sqrt{6^{2}} + \sqrt{6} \cdot 3 - 3 \sqrt{6} - 3 \cdot 3$

Pull terms out from under the radical, assuming positive real numbers.

$6 + \sqrt{6} \cdot 3 - 3 \sqrt{6} - 3 \cdot 3$

Move $3$ to the left of $\sqrt{6}$ .

$6 + 3 \cdot \sqrt{6} - 3 \sqrt{6} - 3 \cdot 3$

Multiply $- 3$ by $3$ .

$6 + 3 \sqrt{6} - 3 \sqrt{6} - 9$

Subtract $9$ from $6$ .

$- 3 + 3 \sqrt{6} - 3 \sqrt{6}$

Subtract $3 \sqrt{6}$ from $3 \sqrt{6}$ .

$- 3 + 0$

Add $- 3$ and $0$ .

$- 3$