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

I think I combine some term, but I'm not sure.

Jun 30, 2020

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$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

just plug in the values and solve!

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

$$=2+3+6+2\sqrt{6}+2\sqrt{18}+2\sqrt{12}$$

$$=11+2\sqrt{6}+2\sqrt{18}+2\sqrt{12}$$

$$=11+2\sqrt{6}+2\sqrt{9\cdot2}+2\sqrt{3\cdot4}$$

$$=\boxed{11+2\sqrt{6}+6\sqrt{2}+4\sqrt{3}}$$

Jun 30, 2020