Question: Simplify: $\dfrac{1-x}{1+x} -$$\dfrac{1+x}{1-x} -$$\dfrac{4x}{1+x²} -$$\dfrac{8x}{1+x⁴}$
Solution:
Given,
$= \dfrac{1-x}{1+x} -\dfrac{1+x}{1-x} -\dfrac{4x}{1+x²} -\dfrac{8x}{1+x⁴}$
$= \dfrac{(1-x)(1-x) - (1+x)(1+x)}{(1+x)(1-x)} -\dfrac{4x}{1+x²} -\dfrac{8x}{1+x⁴}$
$= \dfrac{1 +x² -2x -(1+x² +2x)}{1-x²} -\dfrac{4x}{1+x²} -\dfrac{8x}{1+x⁴}$
$= \dfrac{1 +x² -2x -1 -1 -x² -2x}{1-x²} -\dfrac{4x}{1+x²} -\dfrac{8x}{1+x⁴}$
$= \dfrac{-4x}{1-x²} -\dfrac{4x}{1+x²} -\dfrac{8x}{1+x⁴}$
$= \dfrac{-4x(1+x²) -4x(1-x²)}{(1-x²)(1+x²)} - \dfrac{8x}{1+x⁴}$
$= \dfrac{-4x -4x³ -4x +4x³}{1-x⁴} -\dfrac{8x}{1+x⁴}$
$= \dfrac{-8x}{1-x⁴} -\dfrac{8x}{1 +x⁴}$
$= \dfrac{-8x(1+x⁴) -8x(1-x⁴)}{(1-x⁴)(1+x⁴)}$
$= \dfrac{-8x -8x⁵ -(8x -8x⁵)}{1- x⁸}$
$= \dfrac{-8x -8x⁵ -8x +8x⁵}{1 -x⁸}$
$= \dfrac{-16x}{1 -x⁸}$
$= \dfrac{16x}{x⁸-1}$
= Answer
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