Question: Simplify: $\dfrac{1}{a+1} -$$\dfrac{a}{a²-1} +$$\dfrac{a²}{a⁴-1} +$$\dfrac{a⁴}{a⁸-1}$
Solution:
Given,
$= \dfrac{1}{a+1} -\dfrac{a}{a²-1} +\dfrac{a²}{a⁴-1} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{1}{a+1} -\dfrac{a}{(a+1)(a-1)} +\dfrac{a²}{a⁴-1} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{1(a-1) -a}{(a+1)(a-1)} +\dfrac{a²}{a⁴-1} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{a-1-a}{a² -1} +\dfrac{a²}{(a²+1)(a²-1)} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{-1}{a² -1} +\dfrac{a²}{(a²+1)(a²-1)} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{-1(a²+1) +a²}{(a²-1)(a²+1)} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{-a² -1+a²}{a⁴-1} +\dfrac{a⁴}{a⁸-1}$
$= \dfrac{-1}{a⁴ -1} +\dfrac{a⁴}{(a⁴+1)(a⁴-1)}$
$= \dfrac{-1(a⁴+1) +a⁴}{(a⁴-1)(a⁴+1)}$
$= \dfrac{-a⁴ -1+a⁴}{a⁸-1}$
$= \dfrac{-1}{a⁸-1}$
$= \dfrac{1}{1-a⁸}$
= Answer
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