Question

(3√6+2√8-√32)*√2-√108

Answer 1

3√6*√2+2√8*√2-√32*√2-√108=6√3+16-8-2√28=6√3+8-4√7

Answer 2

Let's simplify the given expression step by step:

$(3\sqrt{6} + 2\sqrt{8} - \sqrt{32}) \times \sqrt{2} - \sqrt{108}$

First, let's simplify the square roots:

$\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$

$\sqrt{8} = \sqrt{2 \times 4} = \sqrt{2} \times \sqrt{4} = 2\sqrt{2}$

$\sqrt{32} = \sqrt{2 \times 16} = \sqrt{2} \times \sqrt{16} = 4\sqrt{2}$

Now the expression becomes:

$(3\sqrt{2} + 2(2\sqrt{2}) - 4\sqrt{2}) \times \sqrt{2} - \sqrt{108}$

Simplify inside the parentheses:

$(3\sqrt{2} + 4\sqrt{2} - 4\sqrt{2}) \times \sqrt{2} - \sqrt{108}$

$(3\sqrt{2}) \times \sqrt{2} - \sqrt{108}$

Now, apply the multiplication:

$3\sqrt{4} - \sqrt{108}$

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

$6 - 6\sqrt{3}$