Question

Evaluate $(\sqrt{3}+\sqrt{2}{)}^6-(\sqrt{3}-\sqrt{2}{)}^6$.

Answer 1

We have $(\sqrt{3}+\sqrt{2}{)}^6-(\sqrt{3}-\sqrt{2}{)}^6$

$=\left[6_{C_0}\right(\sqrt{3}{)}^6+6_{C_1}(\sqrt{3}{)}^5.\sqrt{2}+6_{C_2}(\sqrt{3}{)}^4(\sqrt{2}{)}^2+6_{C_3}(\sqrt{3}{)}^3(\sqrt{2}{)}^3+6_{C_4}(\sqrt{3}{)}^2(\sqrt{2}{)}^4+6_{C_5}(\sqrt{3}\left)\right(\sqrt{2}{)}^5+6_{C_6}\left(\sqrt{2}{)}^6\right]-\left[6_{C_0}\right(\sqrt{3}{)}^6-6_{C_1}(\sqrt{3}{)}^5\sqrt{2}+6_{C_2}(\sqrt{3}{)}^4(\sqrt{2}{)}^2-6_{C_3}(\sqrt{3}{)}^3(\sqrt{2}{)}^3+6_{C_4}(\sqrt{3}{)}^2(\sqrt{2}{)}^4-6_{C_5}(\sqrt{3}\left)\right(\sqrt{2}{)}^5+6_{C_6}\left(\sqrt{2}{)}^6\right]$

$=6_{C_0}(\sqrt{3}{)}^6+6_{C_1}(\sqrt{3}{)}^5\sqrt{2}+6_{C_2}\left(\sqrt{3}{)}^4\right(\sqrt{2}{)}^2+6_{C_3}\left(\sqrt{3}{)}^3\right(\sqrt{2}{)}^3+6_{C_4}\left(\sqrt{3}{)}^2\right(\sqrt{2}{)}^4+6_{C_5}\left(\sqrt{3}\right)(\sqrt{2}{)}^5+6_{C_6}(\sqrt{2}{)}^6-6_{C_0}(\sqrt{3}{)}^6+6_{C_1}(\sqrt{3}{)}^5\sqrt{2}-6_{C_2}\left(\sqrt{3}{)}^4\right(\sqrt{2}{)}^2+6_{C_3}\left(\sqrt{3}{)}^3\right(\sqrt{2}{)}^3-6_{C_4}\left(\sqrt{3}{)}^2\right(\sqrt{2}{)}^4+6_{C_5}\left(\sqrt{3}\right)(\sqrt{2}{)}^5-6_{C_6}(\sqrt{2}{)}^6$

$=2\times 6_{C_1}(\sqrt{3}{)}^5.\sqrt{2}+2\times 6_{C_3}(\sqrt{3}{)}^3(\sqrt{2}{)}^3+2\times 6_{C_5}(\sqrt{3}\left)\right(\sqrt{2}{)}^5$

$=(2\times 6\times 9\sqrt{3}\times \sqrt{2})+(2\times 20\times 3\sqrt{3}\times 2\sqrt{3})+(2\times 6\times \sqrt{3}\times 4\sqrt{2})$

$=108\sqrt{6}+240\sqrt{6}+48\sqrt{6}=396\sqrt{6}$