Question : Simplify the expression:
$\frac{1}{8}\left[\frac{1}{b-1}-\frac{1}{b+1}-\frac{2}{b^2+1}-\frac{4}{b^4+1}\right]$
Option 1: $\frac{1}{b^8-1}$
Option 2: $\frac{8}{b^8+1}$
Option 3: $\frac{8}{b^8-1}$
Option 4: $\frac{1}{b^8+1}$
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Correct Answer: $\frac{1}{b^8-1}$
Solution :
$\frac{1}{8}[\frac{1}{b-1}-\frac{1}{b+1}-\frac{2}{b^2+1}-\frac{4}{b^4+1}]$
= $\frac{1}{8}[\frac{(b+1)-(b-1)}{b^2-1}-\frac{2}{b^2+1}-\frac{4}{b^4+1}]$
= $\frac{1}{8}[\frac{2}{b^2-1}-\frac{2}{b^2+1}-\frac{4}{b^4+1}]$
= $\frac{1}{8}[\frac{2(b^2+1)-2(b^2-1)}{b^4-1}-\frac{4}{b^4+1}]$
= $\frac{1}{8}[\frac{4}{b^4-1}-\frac{4}{b^4+1}]$
= $\frac{1}{8}[\frac{4(b^4+1)-4(b^4-1)}{b^8-1}]$
= $\frac{1}{8}[\frac{8}{b^8-1}]$
= $\frac{1}{b^8-1}$
Hence, the correct answer is $\frac{1}{b^8-1}$.
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