$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ ની કિંમત શોધો.

Firstly, the expression $(a+b)^{6}-(a-b)^{6}$ is simplified by using Binomial Theorem.This can be done as

${(a + b)^6} = {\,^6}{C_0}{a^6} + {\,^6}{C_1}{a^5}b + {\,^6}{C_2}{a^4}{b^2} + {\,^6}{C_3}{a^3}{b^3} + {\,^6}{C_4}{a^2}{b^4} + {\,^6}{C_5}{a^1}{b^5} + {\,^6}{C_6}{b^6}$

$=a^{6}+6 a^{5} b+15 a^{4} b^{2}+20 a^{3} b^{3}+15 a^{2} b^{4}+6 a b^{5}+b^{6}$

${(a - b)^6} = {\,^6}{C_0}{a^6} - {\,^6}{C_1}{a^5}b + {\,^6}{C_2}{a^4}{b^2} - {\,^6}{C_3}{a^3}{b^3} + {\,^6}{C_4}{a^2}{b^4} - {\,^6}{C_5}{a^1}{b^5} + {\,^6}{C_6}{b^6}$

$=a^{6}-6 a^{5} b+15 a^{4} b^{2}-20 a^{3} b^{3}+15 a^{2} b^{4}-6 a b^{5}+b^{6}$

$\therefore(a+b)^{6}-(a-b)^{6}=2\left[6 a^{5} b+20 a^{3} b^{3}+6 a b^{5}\right]$

Putting $a=\sqrt{3}$ and $b=\sqrt{2},$ we obtain

$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}=2\left[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}\right]$

$=2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}]$

$=2 \times 198 \sqrt{6}$

$=396 \sqrt{6}$