Answer
$3x-2\sqrt{3xy} + y$
Work Step by Step
RECALL:
$(a-b)^2=a^2-2ab+b^2$
Use the formula above with $a=\sqrt{3x}$ and $b=\sqrt{y}$ to obtain:
$=(\sqrt{3x})^2 - 2\sqrt{3x} \cdot \sqrt{y} + (\sqrt{y})^2$
Use the rules $(\sqrt{a})^2=a, a \ge 0$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}, a, b, \ge 0$ to obtain:
$\\=3x - 2\sqrt{3x(y)} + y
\\=3x-2\sqrt{3xy} + y$
