Simplify square root of 8b^2* square root of 6b^2+3b^2 square root of 3

$\sqrt{8 b^{2}} \cdot \sqrt{6 b^{2}} + 3 b^{2} \sqrt{3}$

Step 1

Simplify each term.

Rewrite $8 b^{2}$ as ${(2 b)}^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$\sqrt{4 (2) b^{2}} \cdot \sqrt{6 b^{2}} + 3 b^{2} \sqrt{3}$

Rewrite $4$ as $2^{2}$ .

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

Move $2$ .

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

Rewrite $2^{2} b^{2}$ as ${(2 b)}^{2}$ .

$\sqrt{{(2 b)}^{2} \cdot 2} \cdot \sqrt{6 b^{2}} + 3 b^{2} \sqrt{3}$

Pull terms out from under the radical.

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

Reorder $6$ and $b^{2}$ .

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

Pull terms out from under the radical.

$2 b \sqrt{2} \cdot (b \sqrt{6}) + 3 b^{2} \sqrt{3}$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$2 (b \cdot b) \sqrt{2} \cdot \sqrt{6} + 3 b^{2} \sqrt{3}$

Multiply $b$ by $b$ .

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

Multiply $2 b^{2} \sqrt{2} \sqrt{6}$ .

Combine using the product rule for radicals.

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

Multiply $6$ by $2$ .

$2 b^{2} \sqrt{12} + 3 b^{2} \sqrt{3}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$2 b^{2} \sqrt{4 (3)} + 3 b^{2} \sqrt{3}$

Rewrite $4$ as $2^{2}$ .

$2 b^{2} \sqrt{2^{2} \cdot 3} + 3 b^{2} \sqrt{3}$

Pull terms out from under the radical.

$2 b^{2} (2 \sqrt{3}) + 3 b^{2} \sqrt{3}$

Multiply $2$ by $2$ .

$4 b^{2} \sqrt{3} + 3 b^{2} \sqrt{3}$

Step 2

Add $4 b^{2} \sqrt{3}$ and $3 b^{2} \sqrt{3}$ .

$7 b^{2} \sqrt{3}$