Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$
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
$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}=2\left[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}\right]$
$=2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}]$
$=2 \times 198 \sqrt{6}$
$=396 \sqrt{6}$
If the second, third and fourth terms in the expansion of $(x+y)^{\mathrm{n}}$ are $135$,$30$ and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to .............
The sum of the rational terms in the binomial expansion of ${\left( {{2^{\frac{1}{2}}} + {3^{\frac{1}{5}}}} \right)^{10}}$ is
Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of two middle terms in the expansion of $(1+x)^{2 n-1}$
The number of integral terms in the expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$ is equal to
The coefficient of the term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is