Question

Find the product of $(\sqrt{2}+\sqrt{3})(\sqrt{2}+\sqrt{5})(\sqrt{2}+\sqrt{7})$

Answer 1

Consider the binomials $(\sqrt{2}+\sqrt{3})$, $(\sqrt{2}+\sqrt{5})$ and $(\sqrt{2}+\sqrt{7})$.

Firstly multiply the first and second term as follows:

$(\sqrt{2}+\sqrt{3})(\sqrt{2}+\sqrt{5})(\sqrt{2}+\sqrt{7})=\left[\right(\sqrt{2}\times \sqrt{2})+(\sqrt{2}\times \sqrt{5})+(\sqrt{3}\times \sqrt{2})+(\sqrt{3}\times \sqrt{5}\left)\right](\sqrt{2}+\sqrt{7})$

$=(2+\sqrt{10}+\sqrt{6}+\sqrt{15})(\sqrt{2}+\sqrt{7})$

Now multiply the resulting expression:

$(2+\sqrt{10}+\sqrt{6}+\sqrt{15})(\sqrt{2}+\sqrt{7})$

$=(2\times \sqrt{2})+(2\times \sqrt{7})+(\sqrt{10}\times \sqrt{2})+(\sqrt{10}\times \sqrt{7}))+(\sqrt{6}\times \sqrt{2})+(\sqrt{6}\times \sqrt{7})$

$+(\sqrt{15}\times \sqrt{2})+(\sqrt{15}\times \sqrt{7})=2\sqrt{2}+2\sqrt{7}+\sqrt{20}+\sqrt{70}+\sqrt{12}+\sqrt{42}$

$+\sqrt{30}+\sqrt{105}=2\sqrt{2}+2\sqrt{7}+2\sqrt{5}+\sqrt{70}$

$+2\sqrt{3}+\sqrt{42}+\sqrt{30}+\sqrt{105}=2\sqrt{2}+2(\sqrt{3}+\sqrt{5}+\sqrt{7})+\sqrt{2}(\sqrt{15}+\sqrt{35}+\sqrt{21})+\sqrt{105}$

Hence, $(\sqrt{2}+\sqrt{3})(\sqrt{2}+\sqrt{5})(\sqrt{2}+\sqrt{7})=2\sqrt{2}+2(\sqrt{3}+\sqrt{5}+\sqrt{7})+\sqrt{2}(\sqrt{15}+\sqrt{35}+\sqrt{21})+\sqrt{105}$.