Question: $1 + \dfrac{b}{a-b} +$$\dfrac{2ab}{a²+b²} -$$\dfrac{a}{a+b} +$$\dfrac{4a³b}{a⁴+b⁴}$

Solution:
Given,

$= 1+\dfrac{b}{a-b} +\dfrac{2ab}{a²+b²} -\dfrac{a}{a+b} +\dfrac{4a³b}{a⁴+b⁴}$

$= 1 +\dfrac{b}{a-b} -\dfrac{a}{a+b} +\dfrac{2ab}{a²+b²} +\dfrac{4a³b}{a⁴+b⁴}$

$= 1 +\dfrac{b(a+b) -a(a-b)}{(a-b)(a+b)} +\dfrac{2ab}{a²+b²} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{1(a+b)(a-b) +ab +b² -(a² -ab)}{a²-b²} +\dfrac{2ab}{a²+b²} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{a² -b² +ab +b² -a² +ab}{a² -b²} +\dfrac{2ab}{a²+b²} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{2ab}{a² -b²} + \dfrac{2ab}{a² +b²} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{2ab(a²+b²) +2ab(a²-b²)}{(a²-b²)(a²+b²)} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{2ab(a² +b²+a² -b²)}{a⁴-b⁴} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{2ab(2a²)}{a⁴ -b⁴} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{4a³b}{a⁴-b⁴} +\dfrac{4a³b}{a⁴+b⁴}$

$= \dfrac{4a³b(a⁴+b⁴) +4a³b(a⁴-b⁴)}{(a⁴-b⁴)(a⁴+b⁴)}$

$= \dfrac{4a³b(a⁴+b⁴+a⁴-b⁴)}{a⁸-b⁸}$

$= \dfrac{4a³b(2a⁴)}{a⁸-b⁸}$

$= \dfrac{8a⁷b}{a⁸-b⁸}$
= Answer

Related Notes and Solutions:

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