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

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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

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

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

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

$=396 \sqrt{6}$

Similar Questions

The sum of the coefficient of $x^{2 / 3}$ and $x^{-2 / 5}$ in the binomial expansion of $\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9$ is :

  • [JEE MAIN 2024]

If the ratio of the fifth term from the begining to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n$ is $\sqrt{6}: 1$, then the third term from the beginning is:

  • [JEE MAIN 2023]

The term independent of $x$ in the expansion of ${\left( {2x + \frac{1}{{3x}}} \right)^6}$ is

The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :

  • [JEE MAIN 2024]

The positive value of $a$ so that the co-efficient of $x^5$ is equal to that of $x^{15}$ in the expansion of ${\left( {{x^2}\,\, + \,\,\frac{a}{{{x^3}}}} \right)^{10}}$ is