Evaluate (sqrt{3}+sqrt{2})^{6}-(sqrt{3}-sqrt{ | vedclass

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(A) Let $a = \sqrt{3}$ and $b = \sqrt{2}$.Using the Binomial Theorem, we have:$(a+b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k$$(a-b)^6 = \sum_{k=0}

Vedclass · Accepted Answer

$396\sqrt{6}$

Vedclass · Answer

(A) Let $a = \sqrt{3}$ and $b = \sqrt{2}$.Using the Binomial Theorem, we have:$(a+b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} b^k$$(a-b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} (-b)^k$Subtracting the two expressions:$(a+b)^6 - (a-b)^6 = 2 [\binom{6}{1} a^5 b + \binom{6}{3} a^3 b^3 + \binom{6}{5} a b^5]$$= 2 [6 a^5 b + 20 a^3 b^3 + 6 a b^5]$Substituting $a = \sqrt{3}$ and $b = \sqrt{2}$:$= 2 [6(9\sqrt{3})(\sqrt{2}) + 20(3\sqrt{3})(2\sqrt{2}) + 6(\sqrt{3})(4\sqrt{2})]$$= 2 [54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}]$$= 2 [198\sqrt{6}]$$= 396\sqrt{6}$

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

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