Question: Simplify: $\dfrac{b}{a+b} -$$\dfrac{b}{a-b} +$$\dfrac{6b²}{a² -b²} -$$\dfrac{8b⁴}{a⁴ -b⁴}$
Solution:
Given,
$= \dfrac{b}{a+b} -\dfrac{b}{a-b} +\dfrac{6b²}{a²-b²} -\dfrac{8b⁴}{a⁴-b⁴}$
$= \dfrac{b(a-b) -b(a+b)}{(a+b)(a-b)} + \dfrac{6b²}{a²-b²} - \dfrac{8b⁴}{a⁴ -b⁴}$
$= \dfrac{ab -b² -ab -b²}{a² -b²} +\dfrac{6b²}{a²-b²} -\dfrac{8b⁴}{a⁴-b⁴}$
$= \dfrac{-2b²}{a² -b²} +\dfrac{6b²}{a²-b²} -\dfrac{8b⁴}{a⁴-b⁴}$
$= \dfrac{-2b² +6b²}{a² -b²} - \dfrac{8b⁴}{a⁴ -b⁴}$
$= \dfrac{4b²}{a² -b²} -\dfrac{8b⁴}{(a² +b²)(a² -b²)}$
$= \dfrac{4b²(a²+b²) -8b⁴}{(a² -b²)(a² +b²)}$
$= \dfrac{4a²b² +4b⁴ -8b⁴}{(a² -b²)(a² +b²)}$
$= \dfrac{4a²b² -4b⁴}{(a² -b²)(a² +b²)}$
$= \dfrac{4b²(a² -b²)}{(a² -b²)(a² +b²)}$
$= \dfrac{4b²}{a² +b²}$
= Answer
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