Question
Show that:$\frac{1}{1+\text{x}^{\text{a}+\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}}=1$
✓
Answer
$\text{LHS}=-\frac{1}{1+\text{x}^{\text{a}+\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}}=1$ Multiplying the numerators and denominators of two terms on $L.H.S.$
by $x^b$ and $x^a$ respectively, we obtain
$\text{LHS}=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^{\text{a}-\text{b}+\text{b}}}+\frac{\text{x}^\text{a}}{\text{x}^{\text{a}}+\text{x}^{\text{b}-\text{a}+\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{b}+\text{x}^\text{a}}$
$=1$
$=\text{RHS}$
by $x^b$ and $x^a$ respectively, we obtain
$\text{LHS}=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^{\text{a}-\text{b}+\text{b}}}+\frac{\text{x}^\text{a}}{\text{x}^{\text{a}}+\text{x}^{\text{b}-\text{a}+\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{b}+\text{x}^\text{a}}$
$=1$
$=\text{RHS}$
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