Question

Plzz solve it
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Answer 1

HOPe it help u sis................

Answer 2

${i}^{2} = - 1 \\ \\ - {i}^{2} = 1 \\ \\ i = \sqrt{ - 1} \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ {x}^{2} - {y}^{2} = (x - y)(x + y) \\ \\ now \\ \\ 3 {x}^{2} - 2 \sqrt{6} x + 3 \\ \\ = {( \sqrt{3}x) }^{2} - 2( \sqrt{3} x)( \sqrt{2} ) + {( \sqrt{2} )}^{2} + 1 \\ \\ = (\sqrt{3} x + \sqrt{2} ) ^{2} - {i}^{2} \\ \\ = ( \sqrt{3} x + \sqrt{2} + i)( \sqrt{3} x + \sqrt{2} - i)$

hope it helps