Find the value of (a^{2}+sqrt{a^{2}-1})^{4}+( | vedclass

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(A) Let $x = a^{2}$ and $y = \sqrt{a^{2}-1}$.Using the binomial expansion,we know that $(x+y)^{4} + (x-y)^{4} = 2(x^{4} + 6x^{2}y^{2} + y^{4})$.Substi

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$2a^{8}+12a^{6}-10a^{4}-4a^{2}+2$

Vedclass · Answer

(A) Let $x = a^{2}$ and $y = \sqrt{a^{2}-1}$.Using the binomial expansion,we know that $(x+y)^{4} + (x-y)^{4} = 2(x^{4} + 6x^{2}y^{2} + y^{4})$.Substituting the values of $x$ and $y$:$= 2[(a^{2})^{4} + 6(a^{2})^{2}(\sqrt{a^{2}-1})^{2} + (\sqrt{a^{2}-1})^{4}]$$= 2[a^{8} + 6a^{4}(a^{2}-1) + (a^{2}-1)^{2}]$$= 2[a^{8} + 6a^{6} - 6a^{4} + (a^{4} - 2a^{2} + 1)]$$= 2[a^{8} + 6a^{6} - 5a^{4} - 2a^{2} + 1]$$= 2a^{8} + 12a^{6} - 10a^{4} - 4a^{2} + 2$.

Find the value of $(a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$.

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