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$\sqrt{8 - \sqrt{3} - \sqrt{2} + \sqrt{6 + 2 ( \sqrt{3} + \sqrt{2} + \sqrt{6}) } }$
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Answer 1

Final Answer : 3

Steps and Understanding:
1) We know that,
${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \\ + 2ab + 2bc + 2ca$

2) Here,
In
(6 + 2(root(3) + root(2) + root(6) )^(1/2)
a = 1
b = √2
c= √3


$\sqrt{6 + 2(\sqrt{2} + \sqrt{3} + \sqrt{6} )} \\ = > \sqrt{ {1}^{2} + { (\sqrt{2} )}^{2} + {( \sqrt{3} )}^{2} + 2(1 \times \sqrt{2} + \sqrt{2} \times \sqrt{3} + \sqrt{3} \times 1)} \\ \\ \sqrt{ ({1 + \sqrt{2} + \sqrt{3} )}^{2} } = > 1 + \sqrt{2} + \sqrt{3}$


3) Then, using this in given expression


$\sqrt{8 - \sqrt{3} - \sqrt{2} + \sqrt{6 + 2( \sqrt{2} + \sqrt{3} + \sqrt{6} )} } \\ \\ = > \sqrt{8 - \sqrt{3} - \sqrt{2} + 1 + \sqrt{2} + \sqrt{3} } \\ = > \sqrt{8 + 1} \\ = > \sqrt{9} \\ = > 3$