Find (a+b)^{4}-(a-b)^{4}. Hence,evaluate (sqr | vedclass

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(A) Using the Binomial Theorem,the expansions are:$(a+b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Using the Binomial Theorem,the expansions are:$(a+b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}b^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$(a-b)^4 = \binom{4}{0}a^4 - \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 - \binom{4}{3}ab^3 + \binom{4}{4}b^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$Subtracting the two expressions:$(a+b)^4 - (a-b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$$= 8a^3b + 8ab^3 = 8ab(a^2 + b^2)$Now,substitute $a = \sqrt{3}$ and $b = \sqrt{2}$:$8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2) = 8\sqrt{6}(3 + 2) = 8\sqrt{6}(5) = 40\sqrt{6}$.

Find $(a+b)^{4}-(a-b)^{4}$. Hence,evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$. (in $\sqrt{6}$)

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