The value of (sqrt{2} + 1)^6 + (sqrt{2} - 1)^ | vedclass

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(B) Using the binomial expansion formula,$(x + a)^n + (x - a)^n = 2[\binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \binom{n}{4}x^{n-4}a^4 + \binom{n}{6}x^

Vedclass · Accepted Answer

$198$

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(B) Using the binomial expansion formula,$(x + a)^n + (x - a)^n = 2[\binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \binom{n}{4}x^{n-4}a^4 + \binom{n}{6}x^{n-6}a^6]$.Here,$n = 6, x = \sqrt{2}, a = 1$.The binomial coefficients are $\binom{6}{0} = 1, \binom{6}{2} = 15, \binom{6}{4} = 15, \binom{6}{6} = 1$.Substituting the values:$(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2[1 \cdot (\sqrt{2})^6 + 15 \cdot (\sqrt{2})^4 \cdot 1^2 + 15 \cdot (\sqrt{2})^2 \cdot 1^4 + 1 \cdot 1^6]$.$= 2[8 + 15(4) + 15(2) + 1]$.$= 2[8 + 60 + 30 + 1]$.$= 2[99] = 198$.

The value of $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$ is

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