Find the Exact Value square root of 6*(2 square root of 3- square root of 2+ square root of 12)

$\sqrt{6} \cdot (2 \sqrt{3} - \sqrt{2} + \sqrt{12})$

Simplify each term.

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

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

Simplify terms.

Add $2 \sqrt{3}$ and $2 \sqrt{3}$ .

$\sqrt{6} \cdot (4 \sqrt{3} - \sqrt{2})$

Apply the distributive property.

$\sqrt{6} (4 \sqrt{3}) + \sqrt{6} (- \sqrt{2})$

Multiply $\sqrt{6} (4 \sqrt{3})$ .

Combine using the product rule for radicals.

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

Multiply $6$ by $3$ .

$4 \sqrt{18} + \sqrt{6} (- \sqrt{2})$

Multiply $\sqrt{6} (- \sqrt{2})$ .

Combine using the product rule for radicals.

$4 \sqrt{18} - \sqrt{6 \cdot 2}$

Multiply $6$ by $2$ .

$4 \sqrt{18} - \sqrt{12}$

Simplify each term.

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$4 \sqrt{9 (2)} - \sqrt{12}$

Rewrite $9$ as $3^{2}$ .

$4 \sqrt{3^{2} \cdot 2} - \sqrt{12}$

Pull terms out from under the radical.

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

Multiply $3$ by $4$ .

$12 \sqrt{2} - \sqrt{12}$

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

Factor $4$ out of $12$ .

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

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

$12 \sqrt{2} - \sqrt{2^{2} \cdot 3}$

Pull terms out from under the radical.

$12 \sqrt{2} - (2 \sqrt{3})$

Multiply $2$ by $- 1$ .

$12 \sqrt{2} - 2 \sqrt{3}$

The result can be shown in multiple forms.

Exact Form:

$12 \sqrt{2} - 2 \sqrt{3}$

Decimal Form:

$13.50646113 \dots$

Find the Exact Value square root of 6*(2 square root of 3- square root of 2+ square root of 12)

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