Find (x+1)^{6}+(x-1)^{6}. Hence or otherwise, | vedclass

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(A) Using the Binomial Theorem,we have:$(x+1)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} = \binom{6}{0}x^6 + \binom{6}{1}x^5 + \binom{6}{2}x^4 + \binom{6

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(A) Using the Binomial Theorem,we have:$(x+1)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} = \binom{6}{0}x^6 + \binom{6}{1}x^5 + \binom{6}{2}x^4 + \binom{6}{3}x^3 + \binom{6}{4}x^2 + \binom{6}{5}x + \binom{6}{6}$$(x-1)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} (-1)^k = \binom{6}{0}x^6 - \binom{6}{1}x^5 + \binom{6}{2}x^4 - \binom{6}{3}x^3 + \binom{6}{4}x^2 - \binom{6}{5}x + \binom{6}{6}$Adding these two expressions,the odd-powered terms cancel out:$(x+1)^6 + (x-1)^6 = 2[\binom{6}{0}x^6 + \binom{6}{2}x^4 + \binom{6}{4}x^2 + \binom{6}{6}]$$= 2[1 \cdot x^6 + 15 \cdot x^4 + 15 \cdot x^2 + 1]$Now,substitute $x = \sqrt{2}$:$(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6 = 2[(\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1]$$= 2[8 + 15(4) + 15(2) + 1]$$= 2[8 + 60 + 30 + 1]$$= 2[99] = 198$

Find $(x+1)^{6}+(x-1)^{6}$. Hence or otherwise,evaluate $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$.

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